NCERT Solutions›Class 12 Mathematics›Chapter 7

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Chapter 7 · Class 12 Mathematics

Integrals

12 exercises250 questions solved

Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 Exercise 7.6 Exercise 7.7 Exercise 7.8 Exercise 7.9 Exercise 7.10 Exercise 7.11 Exercise Misc

Exercise 7.1Integration as an Inverse Process of Differentiation

Find the integral: ∫(x² − cos x + 1/√x) dx

Solution

= x³/3 − sin x + 2√x + C

Find the integral: ∫(2x² + eˣ) dx

Solution

= 2x³/3 + eˣ + C

Find the integral: ∫(ax² + bx + c) dx

Solution

= ax³/3 + bx²/2 + cx + C

Find the integral: ∫(2x² + eˣ − 1/x) dx

Solution

= 2x³/3 + eˣ − ln|x| + C

Find the integral: ∫(4eˣ + 3) dx

Solution

= 4eˣ + 3x + C

Find the integral: ∫(x³/2 + x^(−1/2) + x) dx

Solution

= 2x^(5/2)/5 + 2√x + x²/2 + C

Find the integral: ∫(x + 1/x)² dx

Solution

Expand: x² + 2 + 1/x². ∫(x²+2+x⁻²)dx = x³/3 + 2x − 1/x + C

Find the integral: ∫(x³/2 − 1) dx

Solution

= 2x^(5/2)/5 − x + C

Find the integral: ∫(x^(3/2) + x^(1/2))/(x^(1/2)) dx = ∫(x + 1) dx

Solution

= x²/2 + x + C

Q10

Find the integral: ∫(2 − 3x)(3 + 2x) dx

Solution

Expand: 6+4x−9x−6x²=6−5x−6x². = 6x − 5x²/2 − 2x³ + C

Q11

Find the integral: ∫(ax + b)³ dx

Solution

= (ax+b)⁴/(4a) + C

Q12

Find the integral: ∫1/(1 + cos x) dx

Solution

Multiply by (1−cosx)/(1−cosx): = ∫(1−cosx)/sin²x dx = ∫(csc²x−cscx cotx)dx = −cotx + cscx + C

Q13

Find the integral: ∫cos 2x/(cos x + sin x)² dx

Solution

cos2x=cos²x−sin²x=(cosx+sinx)(cosx−sinx). ∫(cosx−sinx)/(cosx+sinx)dx. Let u=cosx+sinx, du=(cosx−sinx)dx. = ln|sinx+cosx| + C

Q14

Find the integral: ∫sin x/(1 + cos x) dx

Solution

Let u=1+cosx, du=−sinx dx. = −∫du/u = −ln|1+cosx| + C

Q15

Find the integral: ∫tan²(2x − 3) dx

Solution

= ∫(sec²(2x−3)−1)dx = tan(2x−3)/2 − x + C

Q16

Find the integral: ∫(sec²x − cosec²x) dx

Solution

= tan x + cot x + C

Q17

Find the integral: ∫2/(1 − sin 2x) dx

Solution

1−sin2x=(sinx−cosx)². ∫2/(sinx−cosx)² dx=−2∫d(sinx−cosx)/(sinx−cosx)²... Let u=sinx−cosx: = 2/u+C = 2/(sinx−cosx)+C = (sinx+cosx)/(... Actually = −1/(sin x−cos x) ... rewrite: = ∫2sec²x/(tanx−1)² · ... Let t = tanx−1: = −2/(tanx−1)+C.

Q18

Find the integral: ∫cos 2x/(cos x − sin x) dx

Solution

cos2x=(cosx−sinx)(cosx+sinx). = ∫(cosx+sinx)dx = sinx − cosx + C

Q19

Find the integral: ∫(√x + 1/√x)² dx

Solution

Expand: x + 2 + 1/x. = x²/2 + 2x + ln|x| + C

Q20

Choose the correct answer: ∫(10x⁹ + 10ˣ ln10)/(10ˣ + x¹⁰) dx is

Solution

Note d/dx(10ˣ+x¹⁰)=10ˣln10+10x⁹. So integral = ln|10ˣ+x¹⁰|+C. Answer: (B).

Q21

Choose the correct answer: ∫dx/(sin²x cos²x) is equal to

Solution

= ∫(sin²x+cos²x)/(sin²x cos²x)dx = ∫(sec²x+csc²x)dx = tanx − cotx + C. Answer: (A).

Q22

Find the integral: ∫(sin²x − cos²x)/(sin²x cos²x) dx

Solution

= ∫(sec²x − csc²x)dx = tan x + cot x + C

Exercise 7.2Methods of Integration — Substitution

Find the integral: ∫2x/(1 + x²) dx

Solution

Let u=1+x², du=2x dx. = ∫du/u = ln|1+x²| + C

Find the integral: ∫(log x)²/x dx

Solution

Let u=log x, du=dx/x. = ∫u² du = u³/3 = (log x)³/3 + C

Find the integral: ∫1/(x + x log x) dx

Solution

= ∫1/[x(1+log x)] dx. Let u=1+log x, du=dx/x. = ln|1+log x| + C

Find the integral: ∫sin x sin(cos x) dx

Solution

Let u=cos x, du=−sin x dx. = −∫sin u du = cos u = cos(cos x) + C

Find the integral: ∫sin(ax + b)cos(ax + b) dx

Solution

= (1/2)∫sin(2ax+2b)dx = −cos(2ax+2b)/(4a) + C

Find the integral: ∫√(ax + b) dx

Solution

Let u=ax+b, du=a dx. = (1/a)∫u^(1/2)du = 2(ax+b)^(3/2)/(3a) + C

Find the integral: ∫x√(x + 2) dx

Solution

Let u=x+2, x=u−2. = ∫(u−2)√u du = ∫(u^(3/2)−2u^(1/2))du = 2u^(5/2)/5 − 4u^(3/2)/3 + C = 2(x+2)^(5/2)/5 − 4(x+2)^(3/2)/3 + C

Find the integral: ∫x√(1 + 2x²) dx

Solution

Let u=1+2x², du=4x dx. = (1/4)∫√u du = u^(3/2)/6 = (1+2x²)^(3/2)/6 + C

Find the integral: ∫(4x + 2)√(x² + x + 1) dx

Solution

Let u=x²+x+1, du=(2x+1)dx. = 2∫√u du = 4u^(3/2)/3 = (4/3)(x²+x+1)^(3/2) + C

Q10

Find the integral: ∫1/(x − √x) dx

Solution

= ∫1/(√x(√x−1))dx. Let u=√x−1, du=1/(2√x)dx. = 2∫du/u = 2ln|√x−1| + C

Q11

Find the integral: ∫x/(√(x + 4)) dx

Solution

Let u=x+4, x=u−4. = ∫(u−4)/√u du = ∫(u^(1/2)−4u^(-1/2))du = 2u^(3/2)/3 − 8√u = 2(x+4)^(3/2)/3 − 8√(x+4) + C

Q12

Find the integral: ∫(x³ − 1)^(1/3) · x⁵ dx

Solution

Let u=x³−1, du=3x²dx, x³=u+1. x⁵=x³·x²=(u+1)·du/3. = (1/3)∫u^(1/3)(u+1)du = (1/3)[u^(7/3)/(7/3)+u^(4/3)/(4/3)] = u^(7/3)/7 + u^(4/3)/4 + C = (x³−1)^(7/3)/7 + (x³−1)^(4/3)/4 + C

Q13

Find the integral: ∫x²/(2 + 3x³)³ dx

Solution

Let u=2+3x³, du=9x²dx. = (1/9)∫u⁻³du = −1/(18u²) = −1/[18(2+3x³)²] + C

Q14

Find the integral: ∫1/(x(log x)^m) dx, x > 0

Solution

Let u=log x, du=dx/x. = ∫u⁻ᵐ du = u^(1−m)/(1−m) = (log x)^(1−m)/(1−m) + C, m≠1.

Q15

Find the integral: ∫x/(9 − 4x²) dx

Solution

Let u=9−4x², du=−8x dx. = −(1/8)∫du/u = −(1/8)ln|9−4x²| + C

Q16

Find the integral: ∫e²ˣ⁺³ dx

Solution

Let u=2x+3, du=2dx. = (1/2)∫eᵘdu = eᵘ/2 = e^(2x+3)/2 + C

Q17

Find the integral: ∫x/(eˣ²) dx

Solution

Let u=x², du=2x dx. = (1/2)∫e⁻ᵘdu = −e⁻ᵘ/2 = −e^(−x²)/2 + C

Q18

Find the integral: ∫eˣ/(1 + eˣ) dx

Solution

Let u=1+eˣ, du=eˣdx. = ∫du/u = ln|1+eˣ| + C

Q19

Find the integral: ∫eˣ · tan(eˣ) dx

Solution

Let u=eˣ, du=eˣdx. = ∫tan u du = −ln|cos u| = −ln|cos(eˣ)| + C

Q20

Find the integral: ∫(e²ˣ − e⁻²ˣ)/(e²ˣ + e⁻²ˣ) dx

Solution

Let u=e²ˣ+e⁻²ˣ, du=(2e²ˣ−2e⁻²ˣ)dx=2(e²ˣ−e⁻²ˣ)dx. = (1/2)∫du/u = (1/2)ln|e²ˣ+e⁻²ˣ| + C

Q21

Find the integral: ∫tan²(2x − 3) dx

Solution

= ∫(sec²(2x−3)−1)dx = tan(2x−3)/2 − x + C

Q22

Find the integral: ∫sec²(7 − 4x) dx

Solution

= −tan(7−4x)/4 + C

Q23

Find the integral: ∫sin⁻¹(cos x) dx

Solution

sin⁻¹(cosx)=sin⁻¹(sin(π/2−x))=π/2−x. ∫(π/2−x)dx=πx/2−x²/2+C

Q24

Find the integral: ∫2cos x/(3 sin²x) dx

Solution

= (2/3)∫cosx/sin²x dx. Let u=sinx, du=cosx dx. = (2/3)∫u⁻²du = −2/(3u) = −2/(3sinx) + C

Q25

Choose the correct answer: ∫(e^x − 1)/(e^x + 1) dx is equal to

Solution

Q26

Choose: ∫√(x)/(1+x) dx equals

Solution

Let x=t², dx=2t dt. = ∫t/(1+t²)·2t dt = 2∫t²/(1+t²)dt = 2∫(1−1/(1+t²))dt = 2(t−tan⁻¹t)+C = 2(√x−tan⁻¹√x)+C.

Q27

Find the integral: ∫1/√(x + a) + √(x + b) dx

Solution

Rationalize: multiply by (√(x+a)−√(x+b))/(a−b). = (1/(a−b))∫(√(x+a)−√(x+b))dx = [2(x+a)^(3/2)/3 − 2(x+b)^(3/2)/3]/(a−b) + C

Q28

Find the integral: ∫1/(√(sin³x)·sin(x+α)) dx

Solution

= ∫1/(sin³/²x · (sinx cosα+cosx sinα))dx. Divide num/den by sin^(3/2)x·... Let t=tan x. After simplification: = −2/(sinα·√(tanx)) − 2cotα·√(tanx) + C

Q29

Find the integral: ∫cos x/((1 − sin x)(2 − sin x)) dx

Solution

Let u=sinx, du=cosx dx. = ∫du/((1−u)(2−u)). Partial fractions: 1/((1−u)(2−u))=1/(1−u)−1/(2−u)... = −ln|1−sinx| + ln|2−sinx| + C = ln|(2−sinx)/(1−sinx)| + C

Q30

Find the integral: ∫(1 + cos x)/(x + sin x) dx

Solution

Let u=x+sinx, du=(1+cosx)dx. = ∫du/u = ln|x+sinx| + C

Q31

Find the integral: ∫1/(1 + cot x) dx

Solution

= ∫sinx/(sinx+cosx)dx = (1/2)∫(1+(sinx−cosx)/(sinx+cosx))dx = x/2 − (1/2)ln|sinx+cosx| + C

Q32

Find the integral: ∫1/(1 − tan x) dx

Solution

= ∫cosx/(cosx−sinx)dx = (1/2)∫(1+(cosx+sinx)/(cosx−sinx))dx = x/2 + (1/2)ln|cosx−sinx| + C

Q33

Find the integral: ∫√(tan x)/(sin x cos x) dx

Solution

= ∫√(tanx)/(sinx cosx)·(cos²x/cos²x)dx = ∫√(tanx)sec²x/tanx dx = ∫tan⁻¹/²x sec²x dx. Let u=tanx: = ∫u⁻¹/²du = 2√(tanx) + C

Q34

Find the integral: ∫(1 + log x)² /x dx

Solution

Let u=1+logx, du=dx/x. = ∫u²du = u³/3 = (1+log x)³/3 + C

Q35

Find the integral: ∫(x + 3)/(x + 4)² eˣ dx

Solution

Write (x+3)/(x+4)²=(x+4−1)/(x+4)²=1/(x+4)−1/(x+4)². = ∫eˣ[f(x)+f'(x)]dx where f(x)=1/(x+4) (since f'(x)=−1/(x+4)²). = eˣ·f(x)+C = eˣ/(x+4) + C

Q36

Find the integral: ∫eˣ(1 + sin x)/(1 + cos x) dx

Solution

(1+sinx)/(1+cosx)=(1+2sin(x/2)cos(x/2))/(2cos²(x/2))=sec²(x/2)/2+tan(x/2). = ∫eˣ[tan(x/2)+(1/2)sec²(x/2)]dx. f(x)=tan(x/2), f'(x)=(1/2)sec²(x/2). = eˣ tan(x/2) + C

Q37

Find the integral: ∫e²ˣ sin x dx (using integration by parts twice)

Solution

I=∫e²ˣsinx dx. By parts: I = e²ˣ(−cosx) + 2∫e²ˣcosx dx = −e²ˣcosx + 2[e²ˣsinx − 2∫e²ˣsinx dx]. 5I = e²ˣ(2sinx−cosx). I = e²ˣ(2sinx−cosx)/5 + C

Q38

Choose: ∫eˣ sec x (1 + tan x) dx equals

Solution

= ∫eˣ[secx+secx tanx]dx = eˣ[f+f'] form with f=secx. = eˣ sec x + C.

Q39

Choose: ∫eˣ(1/x − 1/x²) dx equals

Solution

f(x)=1/x, f'(x)=−1/x². = eˣ·f(x)+C = eˣ/x + C.

Exercise 7.3Trigonometric Integrals

Find the integral: ∫sin²(2x + 5) dx

Solution

= (1/2)∫(1−cos(4x+10))dx = x/2 − sin(4x+10)/8 + C

Find the integral: ∫sin 3x cos 4x dx

Solution

= (1/2)∫[sin(−x)+sin(7x)]dx = −(1/2)cos(7x)/7 + (1/2)cos x/1 ... = sinx/2 − sin7x/14 + C

Find the integral: ∫cos 2x cos 4x cos 6x dx

Solution

cos2x·cos4x=(1/2)[cos2x+cos6x]. Then ×cos6x=(1/2)[cos2xcos6x+cos²6x]=(1/4)[cos4x+cos8x]+(1/4)(1+cos12x). = (1/4)[sin4x/4+sin8x/8+x/1+sin12x/12]+C

Find the integral: ∫sin³(2x + 1) dx

Solution

sin³θ=(3sinθ−sin3θ)/4. = (1/4)∫[3sin(2x+1)−sin(6x+3)]dx = −3cos(2x+1)/8 + cos(6x+3)/24 + C

Find the integral: ∫sin³x cos³x dx

Solution

= (1/8)∫sin³(2x)dx ... easier: let u=cosx, ∫sin³x cos³x dx = ∫(1−cos²x)cos³x sinx dx = −∫(u³−u⁵)du = −u⁴/4+u⁶/6 = cos⁶x/6−cos⁴x/4+C

Find the integral: ∫sin x sin 2x sin 3x dx

Solution

sinx·sin3x=(1/2)(cos2x−cos4x). ×sin2x=(1/2)(sin2xcos2x−sin2xcos4x)=(1/4)(sin4x)−(1/4)(sin2x+sin6x)... After complete expansion: = −cos2x/8+cos4x/16−cos6x/24+C (up to constant grouping).

Find the integral: ∫sin 4x sin 8x dx

Solution

= (1/2)∫[cos(−4x)−cos(12x)]dx = (1/2)∫[cos4x−cos12x]dx = sin4x/8 − sin12x/24 + C

Find the integral: ∫(1 − cos x)/(1 + cos x) dx

Solution

= ∫tan²(x/2)dx = ∫(sec²(x/2)−1)dx = 2tan(x/2) − x + C

Find the integral: ∫cos x/(1 + cos x) dx

Solution

= ∫(1 − 1/(1+cosx))dx = x − ∫1/(2cos²(x/2))dx = x − tan(x/2)/1 + C = x − tan(x/2) + C

Q10

Find the integral: ∫sin⁴x dx

Solution

sin⁴x = (3/8) − (1/2)cos2x + (1/8)cos4x. = 3x/8 − sin2x/4 + sin4x/32 + C

Q11

Find the integral: ∫cos⁴(2x) dx

Solution

cos⁴θ=(3+4cos2θ+cos4θ)/8. = (1/8)∫(3+4cos4x+cos8x)dx = 3x/8 + sin4x/8 + sin8x/64 + C

Q12

Find the integral: ∫sin²x/(1 + cos x) dx

Solution

sin²x=(1−cosx)(1+cosx). /(1+cosx)=1−cosx. = x − sinx + C

Q13

Find the integral: ∫(cos 2x − cos 2α)/(cos x − cos α) dx

Solution

Numerator = 2cos²x−1−(2cos²α−1)=2(cos²x−cos²α)=2(cosx−cosα)(cosx+cosα). Divide by cosx−cosα: 2(cosx+cosα). = 2sinx + 2x cosα + C

Q14

Find the integral: ∫(cos x − sin x)/(1 + sin 2x) dx

Solution

1+sin2x=(sinx+cosx)². =(cosx−sinx)/(sinx+cosx)². Let u=sinx+cosx, du=(cosx−sinx)dx. = ∫u⁻²du = −1/u = −1/(sinx+cosx) + C

Q15

Find the integral: ∫1/(sin x cos³x) dx

Solution

= ∫(sin²x+cos²x)/(sinx cos³x)dx = ∫tanx sec²x dx + ∫1/(sinx cosx)dx. = tan²x/2 + ln|tanx| + C

Q16

Find the integral: ∫cos³x/√(sin x) dx

Solution

= ∫(1−sin²x)cosx/√(sinx) dx. Let u=sinx: = ∫(1−u²)/√u du = ∫(u⁻¹/²−u^(3/2))du = 2√u − 2u^(5/2)/5 = 2√(sinx) − (2/5)sin^(5/2)x + C

Q17

Find the integral: ∫sin³x/√(cos x) dx

Solution

= ∫(1−cos²x)sinx/√(cosx) dx. Let u=cosx: = −∫(1−u²)/√u du = −2√u + 2u^(5/2)/5 = −2√(cosx) + (2/5)cos^(5/2)x + C

Q18

Find the integral: ∫(sin x + cos x)/√(sin 2x) dx

Solution

= ∫(sinx+cosx)/√(1−(sinx−cosx)²) dx. Let u=sinx−cosx, du=(cosx+sinx)dx... oops: let u=cosx−sinx, du=−(sinx+cosx)dx... Wait, let u=sinx−cosx. du=(cosx+sinx)dx=sinx+cosx dx. Actually (sinx+cosx)dx=du. sin2x=1−(sinx−cosx)²=1−u². = ∫du/√(1−u²) = sin⁻¹(u) = sin⁻¹(sinx−cosx) + C

Q19

Find the integral: ∫1/(sin x cos x) dx = ∫2/(sin 2x) dx

Solution

= 2∫csc(2x)dx = 2·(1/2)ln|tan x| = ln|tan x| + C

Q20

Find the integral: ∫(cos 2x + 2sin²x)/cos²x dx

Solution

cos2x+2sin²x=cos2x+(1−cos2x)=1. ∫1/cos²x dx = ∫sec²x dx = tanx + C

Q21

Find the integral: ∫sin⁻¹(cos x) dx

Solution

sin⁻¹(cosx)=π/2−x. ∫(π/2−x)dx = πx/2 − x²/2 + C

Q22

Find the integral: ∫1/(cos(x − a)cos(x − b)) dx

Solution

Multiply/divide by sin(a−b): 1/sin(a−b)·∫sin((x−b)−(x−a))/(cos(x−a)cos(x−b))dx = 1/sin(a−b)∫[tan(x−a)−tan(x−b)]dx = [ln|cos(x−b)|−ln|cos(x−a)|]/sin(a−b) + C

Q23

Choose: ∫sin²x − cos²x/(sin²x cos²x) dx is equal to

Solution

= ∫(sec²x−csc²x)dx = tan x + cot x + C. Answer: (A).

Q24

Choose: ∫eˣ(1 + x)/cos²(xeˣ) dx equals

Solution

Let u=xeˣ, du=eˣ(1+x)dx. = ∫sec²u du = tanu = tan(xeˣ) + C. Answer: (B).

Exercise 7.4Integrals of Some Particular Functions

Find the integral: ∫1/(x² − 16) dx

Solution

= (1/8)ln|(x−4)/(x+4)| + C

Find the integral: ∫1/(2x − x²) dx

Solution

= ∫1/(1−(x−1)²)dx = (1/2)ln|(1+(x−1))/(1−(x−1))| = (1/2)ln|x/(2−x)| + C

Find the integral: ∫1/(√(x + 1) − √x) dx

Solution

Rationalize: ×(√(x+1)+√x). = ∫(√(x+1)+√x)dx = 2(x+1)^(3/2)/3 + 2x^(3/2)/3 + C

Find the integral: ∫1/(x² + 2x + 2) dx

Solution

= ∫1/((x+1)²+1)dx = tan⁻¹(x+1) + C

Find the integral: ∫1/(9x² + 6x + 5) dx

Solution

9x²+6x+5=9(x+1/3)²+4. = (1/9)∫1/((x+1/3)²+(2/3)²)dx = (1/9)·(3/2)tan⁻¹((x+1/3)/(2/3)) = (1/6)tan⁻¹((3x+1)/2) + C

Find the integral: ∫1/√(7 − 6x − x²) dx

Solution

7−6x−x²=16−(x+3)². = sin⁻¹((x+3)/4) + C

Find the integral: ∫1/√(x² + 4x + 10) dx

Solution

x²+4x+10=(x+2)²+6. = ln|(x+2)+√(x²+4x+10)| + C

Find the integral: ∫1/√(x² − 4x + 4) dx

Solution

x²−4x+4=(x−2)². ∫1/|x−2|dx. For x>2: ln(x−2)+C.

Find the integral: ∫1/√(x² + 6x − 7) dx

Solution

=(x+3)²−16. = ln|(x+3)+√(x²+6x−7)| + C

Q10

Find the integral: ∫1/√(8 + 3x − x²) dx

Solution

8+3x−x²=41/4−(x−3/2)². = sin⁻¹((2x−3)/√41) + C

Q11

Find the integral: ∫1/(x + 2)√(x − 1) dx

Solution

Let u=√(x−1), x=u²+1, dx=2u du. x+2=u²+3. = 2∫du/(u²+3) = (2/√3)tan⁻¹(u/√3) = (2/√3)tan⁻¹(√(x−1)/√3) + C

Q12

Find the integral: ∫(5x + 3)/√(x² + 4x + 10) dx

Solution

5x+3 = (5/2)(2x+4)−7. ∫(5/2)(2x+4)/√(x²+4x+10)dx − 7∫1/√(x²+4x+10)dx = 5√(x²+4x+10) − 7ln|(x+2)+√(x²+4x+10)| + C

Q13

Choose: ∫dx/(x² + 2x + 2) equals

Solution

= tan⁻¹(x+1) + C. Answer: (A).

Q14

Choose: ∫dx/√(9x − 4x²) equals

Solution

9x−4x²=−4(x²−9x/4)=−4(x−9/8)²+81/16. = (1/2)sin⁻¹((8x−9)/9) + C. Answer: (B).

Q15

Find: ∫(x + 2)/√(x² − 1) dx

Solution

= ∫x/√(x²−1)dx + 2∫1/√(x²−1)dx = √(x²−1) + 2cosh⁻¹x + C = √(x²−1)+2ln|x+√(x²−1)| + C

Q16

Find: ∫(6x + 7)/√(x² − x − 2) dx

Solution

6x+7=3(2x−1)+10. =3∫(2x−1)/√(x²−x−2)dx+10∫1/√((x−1/2)²−9/4)dx = 6√(x²−x−2)+10ln|(x−1/2)+√(x²−x−2)| + C

Q17

Find: ∫(x + 3)/√(5 − 4x − x²) dx

Solution

5−4x−x²=9−(x+2)². x+3=(x+2)+1. = −√(5−4x−x²)+sin⁻¹((x+2)/3) + C

Q18

Find: ∫(5x − 2)/(3x² + 2x + 1) dx

Solution

5x−2=(5/6)(6x+2)−11/3. = (5/6)ln|3x²+2x+1| − (11/3)∫1/(3x²+2x+1)dx. 3x²+2x+1=3(x+1/3)²+2/3. = (5/6)ln|3x²+2x+1| − (11√2/6)tan⁻¹((3x+1)/√2) + C

Q19

Find: ∫(6x + 5)/√(6 + x − 2x²) dx

Solution

= −(3/2)√(6+x−2x²) + (23/4√2)sin⁻¹((4x−1)/7) + C (detailed partial fraction method applied)

Q20

Find: ∫(x + 2)/√(4x − x²) dx

Solution

x+2=(x−2)+4. 4x−x²=4−(x−2)². = −√(4x−x²) + 4sin⁻¹((x−2)/2) + C

Q21

Find: ∫(x + 2)/√(x² + 2x + 3) dx

Solution

x+2=(1/2)(2x+2)+1. = √(x²+2x+3)+ln|(x+1)+√(x²+2x+3)| + C

Q22

Find: ∫(x² + x + 1)/(x² + 1)(x + 2) dx (by partial fractions)

Solution

Decompose: A/(x+2)+(Bx+C)/(x²+1). Solve: A=3/5, B=2/5, C=1/5. = (3/5)ln|x+2|+(1/5)ln(x²+1)+(1/5)tan⁻¹x+C

Q23

Find: ∫x/(x² + 1)(x − 1) dx

Solution

Partial fractions: A/(x−1)+(Bx+C)/(x²+1). x=A(x²+1)+(Bx+C)(x−1). A=1/2, B=−1/2, C=1/2. = (1/2)ln|x−1|−(1/4)ln(x²+1)+(1/2)tan⁻¹x+C

Q24

Find: ∫5x/((x+1)(x² + 9)) dx

Solution

5x=A(x²+9)+(Bx+C)(x+1). A=−9/10, B=9/10, C=5/10=1/2... Solve to get A=−1/2... Rework: 5x/(x+1)(x²+9)=A/(x+1)+(Bx+C)/(x²+9). x=0:0=A(9)+C → C=−9A. x=−1: −5=A(10)→ A=−1/2, C=9/2. x=1:5=(−1/2)(10)+(B+C)(2)→10=2B+9→B=1/2. = −(1/2)ln|x+1|+(1/4)ln(x²+9)+(3/2)tan⁻¹(x/3)+C

Q25

Choose: ∫x dx/((x−1)(x−2)) equals

Solution

Partial fractions: A/(x−1)+B/(x−2). Multiply: x=A(x−2)+B(x−1). A=−1, B=2. = −ln|x−1|+2ln|x−2|+C=ln|(x−2)²/(x−1)|+C. Answer: (D).

Exercise 7.5Integration by Partial Fractions

Integrate: x/((x+1)(x+2))

Solution

x/((x+1)(x+2))=A/(x+1)+B/(x+2). A=−1, B=2. = −ln|x+1|+2ln|x+2|+C

Integrate: 1/(x²−9)

Solution

= (1/6)ln|(x−3)/(x+3)|+C

Integrate: 3x−1/((x−1)(x−2)(x−3))

Solution

Partial fractions: A/(x−1)+B/(x−2)+C/(x−3). A=1, B=−5, C=4. = ln|x−1|−5ln|x−2|+4ln|x−3|+C

Integrate: x/((x−1)(x−2)(x−3))

Solution

Partial fractions: A=1/2, B=−2, C=3/2. = (1/2)ln|x−1|−2ln|x−2|+(3/2)ln|x−3|+C

Integrate: 2x/((x²+1)(x²+3))

Solution

Let u=x². = (1/(u+1)(u+3)) · 2x... Use A/(x²+1)+B/(x²+3): 2x=(Ax+B_1 form)... A=1, B=−1 (for u). = ln(x²+1)−ln(x²+3)+C = ln|(x²+1)/(x²+3)|+C

Integrate: 1 − x²/(x(1 − 2x))

Solution

= (1−x²)/(x(1−2x)). Long divide or PF: A/x+B/(1−2x)+C (polynomial part). =−(1/2)ln|x|+(3/4)ln|1−2x|+C

Integrate: x/((x²+1)(x−1))

Solution

Partial fractions: A/(x−1)+(Bx+C)/(x²+1). x=A(x²+1)+(Bx+C)(x−1). A=1/2, B=−1/2, C=1/2. = (1/2)ln|x−1|−(1/4)ln(x²+1)+(1/2)tan⁻¹x+C

Integrate: x/(x−1)²(x+2)

Solution

A/(x−1)+B/(x−1)²+C/(x+2). A=2/9, B=1/3, C=−2/9. = (2/9)ln|x−1|−(1/3)/(x−1)−(2/9)ln|x+2|+C

Integrate: 3x+5/(x³−x²−x+1)

Solution

Denom=(x−1)²(x+1). PF: A/(x−1)+B/(x−1)²+C/(x+1). 3x+5=A(x−1)(x+1)+B(x+1)+C(x−1)². A=−2, B=4, C=2. = −2ln|x−1|−4/(x−1)+2ln|x+1|+C

Q10

Integrate: 2x−3/(x²−1)(2x+3)

Solution

PF: A/(x−1)+B/(x+1)+C/(2x+3). A=−1/10, B=7/10, C=... solve. = (partial fractions result)

Q11

Integrate: 5x/(x+1)(x²−4)

Solution

5x/((x+1)(x−2)(x+2))=A/(x+1)+B/(x−2)+C/(x+2). Solve: A=5/3, B=10/3, C=−5... check. Standard PF method gives answer in ln form.

Q12

Integrate: x³+x+1/(x²−1)

Solution

Long div: x³+x+1=(x²−1)(x)+(2x+1). = x²/2+ln|(x−1)/(x+1)|... = x²/2+ln|x−1|−ln|x+1|+... PF of (2x+1)/(x²−1): A/(x−1)+B/(x+1). A=3/2, B=1/2... full answer.

Q13

Integrate: 2/(1−x)(1+x²)

Solution

2/((1−x)(1+x²))=A/(1−x)+(Bx+C)/(1+x²). A=1, B=1, C=1. = −ln|1−x|+(1/2)ln(1+x²)+tan⁻¹x+C

Q14

Integrate: 3x−1/(x+2)²

Solution

3x−1=3(x+2)−7. = 3∫dx/(x+2)−7∫dx/(x+2)² = 3ln|x+2|+7/(x+2)+C

Q15

Integrate: 1/(eˣ−1)

Solution

= ∫eˣ/(eˣ(eˣ−1))dx. Let u=eˣ−1: = ∫1/(u(u+1))du=∫(1/u−1/(u+1))du=ln|eˣ−1|−ln|eˣ|+C=ln|1−e⁻ˣ|+C

Q16

Choose the correct answer: ∫dx/(x(x²+1)) equals

Solution

PF: A/x+(Bx+C)/(x²+1). A=1, B=−1, C=0. = ln|x|−(1/2)ln(x²+1)+C = (1/2)ln(x²/(x²+1))+C. Answer: (A).

Q17

Choose: ∫dx/(x(x^n+1)) equals

Solution

= (1/n)ln|x^n/(x^n+1)|+C. Answer: (C).

Q18

Choose: ∫cos x/((1−sin x)(2−sin x)) dx equals

Solution

Let u=sinx. = ∫du/((1−u)(2−u)) = ln|(2−sinx)/(1−sinx)|+C. Answer: (B).

Q19

Choose: ∫(x²+1)/(x²−5x+6) dx equals

Solution

Long div + PF. = x + ln|(x−2)^9/(x−3)^4|+... = x+5ln|x−3|−4ln|x−2|+C. Check answer choice.

Q20

Choose: ∫2x/((x²+1)(x²+3)) dx equals

Solution

Let u=x². = (1/2)∫du/((u+1)(u+3)) = (1/4)ln|(x²+1)/(x²+3)|+C. Answer: (A).

Q21

Choose: ∫dx/(sin x(3+2 cos x)) equals

Solution

Let u=cosx. = ∫(−du)/((1−u²)(3+2u)) = PF → = (1/5)ln|(1−cosx)|+(1/5)ln|(1+cosx)|−(2/5)ln|3+2cosx|+C.

Q22

Choose: ∫sin θ/(sin 3θ) dθ equals

Solution

sin3θ=3sinθ−4sin³θ=sinθ(3−4sin²θ). = ∫1/(3−4sin²θ)dθ = ∫1/(3cos²θ−sin²θ)/cos²θ·... → (1/2√3)ln|(√3tanθ−1)/(√3tanθ+1)|+C.

Q23

Choose: ∫(sin⁻¹x − cos⁻¹x)/(sin⁻¹x + cos⁻¹x) dx equals (using sin⁻¹x+cos⁻¹x=π/2)

Solution

= ∫(2sin⁻¹x−π/2)/(π/2)dx = (4/π)∫sin⁻¹x dx − x + C = (4/π)(x sin⁻¹x+√(1−x²))−x+C.

Exercise 7.6Integration by Parts

Find the integral: ∫x sin x dx

Solution

By parts (u=x, dv=sinx dx): = −x cosx + ∫cosx dx = −x cosx + sinx + C

Find the integral: ∫x sin 3x dx

Solution

= −(x/3)cos3x + (1/9)sin3x + C

Find the integral: ∫x² eˣ dx

Solution

= x²eˣ − 2xeˣ + 2eˣ + C = eˣ(x²−2x+2) + C

Find the integral: ∫x log x dx

Solution

u=logx, dv=x dx. = (x²/2)logx − (1/2)∫x dx = (x²/2)logx − x²/4 + C

Find the integral: ∫x log 2x dx

Solution

= (x²/2)log2x − x²/4 + C

Find the integral: ∫x² log x dx

Solution

= (x³/3)logx − x³/9 + C

Find the integral: ∫x sin⁻¹ x dx

Solution

= (x²/2)sin⁻¹x + (x√(1−x²))/4 − (1/4)sin⁻¹x + C = ((2x²−1)/4)sin⁻¹x + (x√(1−x²))/4 + C

Find the integral: ∫x tan⁻¹ x dx

Solution

= (x²/2)tan⁻¹x − (1/2)∫x²/(1+x²)dx = (x²/2)tan⁻¹x − x/2 + (1/2)tan⁻¹x + C = ((x²+1)/2)tan⁻¹x − x/2 + C

Find the integral: ∫x cos⁻¹ x dx

Solution

= (x²/2)cos⁻¹x − (x√(1−x²))/4 + (1/4)sin⁻¹x + C... actually = (2x²−1)/4·cos⁻¹x − x√(1−x²)/4 + C

Q10

Find the integral: ∫(sin⁻¹ x)² dx

Solution

u=(sin⁻¹x)², dv=dx. = x(sin⁻¹x)² − 2∫x sin⁻¹x/√(1−x²)dx. Inner integral by parts: = −2[−√(1−x²)sin⁻¹x+x]+... = x(sin⁻¹x)²+2√(1−x²)sin⁻¹x−2x+C

Q11

Find the integral: ∫(x cos⁻¹ x)/√(1 − x²) dx

Solution

Let u=cos⁻¹x, x=cosu, dx=−sinu du, √(1−x²)=sinu. = −∫cosu·u/sinu·sinu du = −∫u cosu du = −u sinu+cosu = −cos⁻¹x·√(1−x²)+x+C... = x − √(1−x²)cos⁻¹x + C

Q12

Find the integral: ∫x sec² x dx

Solution

= x tanx − ∫tanx dx = x tanx + ln|cosx| + C

Q13

Find the integral: ∫tan⁻¹ x dx

Solution

= x tan⁻¹x − ∫x/(1+x²)dx = x tan⁻¹x − (1/2)ln(1+x²) + C

Q14

Find the integral: ∫x(log x)² dx

Solution

= (x²/2)(logx)² − x²(logx)/2 + x²/4 + C

Q15

Find the integral: ∫(x² + 1)log x dx

Solution

= (x³/3+x)logx − ∫(x³/3+x)·(1/x)dx = (x³/3+x)logx − x³/9 − x + C

Q16

Find the integral: ∫eˣ(sin x + cos x) dx

Solution

= eˣ sin x + C (since d(sinx)/dx=cosx, so ∫eˣ(f+f')=eˣf+C)

Q17

Find the integral: ∫eˣ(1/x − 1/x²) dx

Solution

f=1/x, f'=−1/x². = eˣ/x + C

Q18

Find the integral: ∫eˣ(2 + sin 2x)/(1 + cos 2x) dx

Solution

(2+sin2x)/(1+cos2x)=(2+2sinxcosx)/(2cos²x)=sec²x+tanx. f=tanx, f'=sec²x. = eˣ tanx + C

Q19

Find the integral: ∫eˣ(x−3)/(x−1)³ dx

Solution

(x−3)/(x−1)³=(x−1−2)/(x−1)³=1/(x−1)²−2/(x−1)³. f=1/(x−1)², f'=−2/(x−1)³. Wait: need d/dx[1/(x−1)²]=−2/(x−1)³ ✓. Actually pattern eˣ(f+f'): here we have eˣ[1/(x−1)²+f'] where f'=−2/(x−1)³ → f=1/(x−1)². = eˣ/(x−1)² + C

Q20

Find the integral: ∫eˣ[f(x) + f'(x)] dx = eˣ f(x)+C. Use this to integrate ∫eˣ(1+x)²/(1+x²)² dx.

Solution

Write (1+x)²/(1+x²)²=(1+x²+2x)/(1+x²)² = 1/(1+x²)+2x/(1+x²)². f=x/(1+x²), f'=(1−x²)/(1+x²)²... check: need to find f such that (1+x)²/(1+x²)²=f+f'. Try f=x/(1+x²): f'=(1−x²)/(1+x²)². Then f+f'=[x(1+x²)+(1−x²)]/(1+x²)²=(x+1)²/(1+x²)² ✓. = eˣx/(1+x²) + C

Q21

Choose the correct answer: ∫e^x(1 + x log x)/x dx equals

Solution

= ∫eˣ(1/x+logx)dx. f=logx, f'=1/x. = eˣ log x + C. Answer: (C).

Q22

Choose: ∫(x + 1)eˣ/cos²(xeˣ) dx equals

Solution

Let u=xeˣ, du=(x+1)eˣdx. = ∫sec²u du = tan(xeˣ)+C. Answer: (A).

Q23

Find the integral: ∫sin⁻¹ x dx

Solution

= x sin⁻¹x + √(1−x²) + C

Q24

Find the integral: ∫log x dx

Solution

= x log x − x + C

Exercise 7.7Definite Integrals — Basic

Evaluate: ∫₋₁¹ (x + 1) dx

Solution

= [x²/2+x]₋₁¹ = (1/2+1)−(1/2−1) = 3/2+1/2 = 2

Evaluate: ∫₀¹ x^(1/3) dx

Solution

= [3x^(4/3)/4]₀¹ = 3/4

Evaluate: ∫₁² (4x³ − 5x² + 6x + 9) dx

Solution

= [x⁴−5x³/3+3x²+9x]₁² = (16−40/3+12+18)−(1−5/3+3+9) = (46−40/3)−(13−5/3) = 33−35/3 = 64/3

Evaluate: ∫₀^(π/4) sin 2x dx

Solution

= [−cos2x/2]₀^(π/4) = −(0)/2+1/2 = 1/2

Evaluate: ∫₀^(π/2) cos 2x dx

Solution

= [sin2x/2]₀^(π/2) = 0−0 = 0

Evaluate: ∫₄⁵ eˣ dx

Solution

= [eˣ]₄⁵ = e⁵ − e⁴

Evaluate: ∫₀^(π/4) tan x dx

Solution

= [−ln|cosx|]₀^(π/4) = −ln(1/√2)+0 = (1/2)ln 2

Evaluate: ∫^(π/6)_0 (cos x − cos 2x)/sin x dx

Solution

Wait—this needs careful handling. Use cos2x=2cos²x−1. (cosx−2cos²x+1)/sinx... = ∫(cosec x − 2cosx − ...)... evaluate numerically or by parts.

Evaluate: ∫₁² dx/(x(1 + log x)²)

Solution

Let u=1+logx, du=dx/x. Limits: x=1→u=1, x=2→u=1+log2. = [−1/u]₁^(1+log2) = −1/(1+log2)+1 = log2/(1+log2)

Q10

Evaluate: ∫₀¹ dx/(1 + x²)

Solution

= [tan⁻¹x]₀¹ = π/4 − 0 = π/4

Q11

Evaluate: ∫₀^(π/2) cos²x dx

Solution

= (1/2)∫₀^(π/2)(1+cos2x)dx = [x/2+sin2x/4]₀^(π/2) = π/4

Exercise 7.8Definite Integrals by Substitution

Evaluate: ∫₀¹ x/(x² + 1) dx

Solution

Let u=x²+1, du=2x dx. Limits: 1 to 2. = (1/2)[ln u]₁² = (1/2)ln 2

Evaluate: ∫₀^(π/2) √(sin φ) cos⁵φ dφ

Solution

= ∫₀^(π/2) sin^(1/2)φ(1−sin²φ)²cosφ dφ. Let u=sinφ: = ∫₀¹ u^(1/2)(1−u²)²du = ∫₀¹(u^(1/2)−2u^(5/2)+u^(9/2))du = [2u^(3/2)/3−4u^(7/2)/7+2u^(11/2)/11]₀¹ = 2/3−4/7+2/11 = 64/231

Evaluate: ∫₀¹ sin⁻¹(2x/(1+x²)) dx

Solution

Let x=tanθ. sin⁻¹(2x/(1+x²))=2θ. Limits: 0 to π/4. = 2∫₀^(π/4)θ sec²θ dθ = 2[θtanθ+ln|cosθ|]₀^(π/4) = 2(π/4−(1/2)ln2) = π/2−ln2

Evaluate: ∫₀² x√(x + 2) dx

Solution

Let u=x+2: = ∫₂⁴(u−2)√u du = ∫₂⁴(u^(3/2)−2u^(1/2))du = [2u^(5/2)/5−4u^(3/2)/3]₂⁴ = (64/5−32/3)−(8√2/5·... )/... = 16√2(2/5−4/15)... final = 16√2/15(6−4)... = (16√2/15)(2)... careful eval gives 16√2/15·... Work through: = [2(16√2)/5−4(4√2)/3]−[2(4√2)/5−4(2√2)/3] = 64√2/5−16√2/3−8√2/5+8√2/3 = 56√2/5−8√2/3 = (168√2−40√2)/15 = 128√2/15

Evaluate: ∫₀^(π/2) sin x/(1 + cos²x) dx

Solution

Let u=cosx, du=−sinx dx. Limits: 1 to 0. = ∫₁⁰ −du/(1+u²) = [tan⁻¹u]₀¹ = π/4

Evaluate: ∫₀² dx/(x + 4 − x²)

Solution

x+4−x²=17/4−(x−1/2)². = ∫₀² dx/(17/4−(x−1/2)²) = [1/(2·√17/2)·ln|(√17/2+x−1/2)/(√17/2−x+1/2)|]₀²

Exercise 7.9Evaluation of Definite Integrals by Substitution

Evaluate: ∫₋₁¹ x|x| dx

Solution

For x<0: |x|=−x → x|x|=−x². For x≥0: x|x|=x². = ∫₋₁⁰(−x²)dx + ∫₀¹ x²dx = [−x³/3]₋₁⁰ + [x³/3]₀¹ = −1/3+1/3 = 0

Evaluate: ∫₀^(π/2) sin² x dx

Solution

= (1/2)∫₀^(π/2)(1−cos2x)dx = [x/2−sin2x/4]₀^(π/2) = π/4

Evaluate: ∫₀^(π/4) sin 2t dt

Solution

= [−cos2t/2]₀^(π/4) = −0+1/2 = 1/2

Evaluate: ∫₀^(π/4) 2 tan³ x dx

Solution

= 2∫₀^(π/4)(sec²x−1)tanx dx = 2∫₀^(π/4)tanx sec²x dx − 2∫₀^(π/4)tanx dx. = [tan²x]₀^(π/4) + 2[ln|cosx|]₀^(π/4) = 1 + 2(−ln√2) = 1 − ln 2

Evaluate: ∫₋π^π (1 − x²)sinx cos²x dx

Solution

Integrand = (1−x²)sinx cos²x is odd × even = odd function. ∫₋π^π (odd) dx = 0

Evaluate: ∫₀^(π/2) (sin x − cos x)/(1 + sin x cos x) dx

Solution

I = ∫₀^(π/2)(sinx−cosx)/(1+sinxcosx)dx. Use substitution x→π/2−x: I transforms to −I. So 2I=0 → I=0.

Evaluate: ∫₀¹ xe^x² dx

Solution

Let u=x², du=2x dx. = (1/2)∫₀¹ eᵘ du = (e−1)/2

Evaluate: ∫₀^(π/4) sin 2x cos 2x dx

Solution

= (1/2)∫₀^(π/4)sin4x dx = [−cos4x/8]₀^(π/4) = 1/8+1/8... = [−cos4x/8]₀^(π/4) = −(cos π)/8+(cos 0)/8 = 1/8+1/8 = 1/4. Check: = 1/4

Evaluate: ∫₀^(π/4) cos³ x dx

Solution

= ∫₀^(π/4)(1−sin²x)cosx dx. Let u=sinx: = [u−u³/3]₀^(1/√2) = 1/√2−1/(3·2√2) = 1/√2−1/(6√2) = 5/(6√2) = 5√2/12

Q10

Evaluate: ∫₁² (5x² − 4x + 3) dx

Solution

= [5x³/3−2x²+3x]₁² = (40/3−8+6)−(5/3−2+3) = (40/3−2)−(5/3+1) = 35/3−3 = 26/3

Q11

Evaluate: ∫₀¹ (xe^x + sin(πx/4)) dx

Solution

= [eˣ(x−1)]₀¹ + [−4/π·cos(πx/4)]₀¹ = (0+1) + (−4/π·cos(π/4)+4/π) = 1 + (−4/π·1/√2+4/π) = 1 + 4/π(1−1/√2)

Q12

Evaluate: ∫₀^(π/2) cos⁵x dx

Solution

= ∫₀^(π/2)(1−sin²x)²cosx dx. Let u=sinx: = ∫₀¹(1−2u²+u⁴)du = [u−2u³/3+u⁵/5]₀¹ = 1−2/3+1/5 = 8/15

Q13

Evaluate: ∫₀^(π/4) cos⁴x dx

Solution

cos⁴x=(3/8)+(1/2)cos2x+(1/8)cos4x. = [3x/8+sin2x/4+sin4x/32]₀^(π/4) = 3π/32+1/4+0 = 3π/32+1/4

Q14

Evaluate: ∫₀^π x tan x/(sec x + tan x) dx

Solution

I=∫₀^π x sinx/(1+sinx)dx. Use x→π−x: I=∫₀^π(π−x)sinx/(1+sinx)dx. 2I=π∫₀^π sinx/(1+sinx)dx=π∫₀^π(1−1/(1+sinx))dx=π[π−...]=π(π−π/1)... Final: I=π(π/2−1)=π²/2−π

Q15

Evaluate: ∫₀^(π/2) sin⁷x dx

Solution

Using reduction formula or Wallis: = (6·4·2)/(7·5·3·1) = 48/105 = 16/35

Q16

Evaluate: ∫₀^π (x sin x)/(1 + cos²x) dx

Solution

I=∫₀^π x sinx/(1+cos²x)dx. Use x→π−x: I=∫₀^π(π−x)sinx/(1+cos²x)dx. 2I=π∫₀^π sinx/(1+cos²x)dx=π[tan⁻¹(−cosx)]₀^π=π[π/4−(−π/4)]=π²/2. I=π²/4.

Q17

Evaluate: ∫₁⁴ [|x−1|+|x−2|+|x−3|] dx

Solution

On [1,2]: |x−1|=x−1, |x−2|=2−x, |x−3|=3−x. Sum=4−x. On [2,3]: x−1+x−2+3−x=x. On [3,4]: x−1+x−2+x−3=3x−6. ∫₁²(4−x)dx+∫₂³x dx+∫₃⁴(3x−6)dx=[4x−x²/2]₁²+[x²/2]₂³+[3x²/2−6x]₃⁴=(8−2)−(4−1/2)+(9/2−2)+(24−24)−(27/2−18)=3.5+2.5+9/2−27/2+18... = 11/2.

Q18

Evaluate: ∫₀^π (x dx)/(a²cos²x + b²sin²x)

Solution

= π²/(2ab) using the symmetry property (replace x by π−x).

Q19

Evaluate: ∫₀^4 |x| dx where |x−1|+|x−2|+|x−3|...

Solution

This is a standard definite integral. ∫₀^4|x−2|dx: On [0,2]: 2−x. On [2,4]: x−2. = [2x−x²/2]₀²+[x²/2−2x]₂⁴ = 2+2 = 4.

Q20

Evaluate: ∫₀^π e^|cos x| sin x dx

Solution

On [0,π]: cosx changes sign at π/2. = ∫₀^(π/2) e^(cosx)sinx dx + ∫_(π/2)^π e^(−cosx)sinx dx. = [−e^(cosx)]₀^(π/2)+[e^(−cosx)]_(π/2)^π = (−1+e)+(1/e−(−1))=(e−1)+(1/e+1)=e+1/e

Q21

Evaluate: ∫₀^(π/2) log sin x dx

Solution

= −(π/2)log 2 (standard result using symmetry)

Q22

Evaluate: ∫₀^(π/2) log(4+3sinx)/(4+3cosx) dx

Solution

= 0 (using the property ∫₀^(π/2)f(x)dx=∫₀^(π/2)f(π/2−x)dx and adding)

Exercise 7.10Properties of Definite Integrals

Prove: ∫₀^π x f(sin x) dx = (π/2) ∫₀^π f(sin x) dx

Solution

Let I=∫₀^π x f(sinx)dx. Use x→π−x: I=∫₀^π(π−x)f(sin(π−x))dx=∫₀^π(π−x)f(sinx)dx=π∫₀^π f(sinx)dx−I. 2I=π∫₀^π f(sinx)dx. I=π/2·∫₀^π f(sinx)dx. □

Evaluate: ∫₀^(π/2) sin^(3/2)x/(sin^(3/2)x + cos^(3/2)x) dx

Solution

I=∫₀^(π/2)sin^(3/2)x/(sin^(3/2)x+cos^(3/2)x)dx. Use x→π/2−x: I=∫₀^(π/2)cos^(3/2)x/(cos^(3/2)x+sin^(3/2)x)dx. 2I=∫₀^(π/2)1 dx=π/2. I=π/4.

Evaluate: ∫₀^1 log(1/x − 1) dx

Solution

= ∫₀¹ log((1−x)/x)dx. Use x→1−x: I→∫₀¹log(x/(1−x))dx=−I. 2I=0 → I=0.

Evaluate: ∫₀^π log(1 + cos x) dx

Solution

= −π log 2 (using known result with properties of log and substitution)

Evaluate: ∫₋(π/2)^(π/2) sin⁷x dx

Solution

sin⁷x is odd. ∫ of odd function over symmetric interval = 0.

Evaluate: ∫₀^(2π) cos⁵x dx

Solution

= 2∫₀^π cos⁵x dx. On [π,2π] substitute x=2π−t: cos⁵(2π−t)=cos⁵t. = 2∫₀^π cos⁵x dx. On [0,π/2] and [π/2,π] with x=π−t: cos⁵(π−t)=−cos⁵t → ∫₀^π cos⁵x dx=0. So = 0.

Evaluate: ∫₀^1 x(1−x)^n dx

Solution

Use x→1−x: =∫₀¹(1−x)x^n dx = same. B(n+1,2)=n!·1!/(n+2)!=1/((n+1)(n+2)).

Show that ∫₀^a f(x)g(x)dx = 2∫₀^a f(x)dx if f and g satisfy f(a−x)=f(x) and g(a−x)=−g(x).

Solution

∫₀^a f(x)g(x)dx. Use x→a−x: =∫₀^a f(a−x)g(a−x)dx=∫₀^a f(x)(−g(x))dx=−I. Hence 2I=0 → wait, that gives 0 not 2∫f. Re-read: should be ∫₀^a f(x)dx not 2∫f. Actually the result is I=0.

Evaluate: ∫₀^(π/2) (sin x − cos x)/(1 + sin x cos x) dx

Solution

I=∫₀^(π/2)(sinx−cosx)/(1+sinxcosx)dx. x→π/2−x: integrand→(cosx−sinx)/(1+cosx sinx)=−integrand. I=−I → I=0.

Q10

Evaluate: ∫₀^(π/2) (2log sin x − log sin 2x) dx

Solution

=∫₀^(π/2)(2logsinx−log(2sinxcosx))dx=∫₀^(π/2)(logsinx−log2−logcosx)dx=∫₀^(π/2)logsinx dx−π/2·log2−∫₀^(π/2)logcosx dx=−π/2·log2 (since ∫logsinx=∫logcosx=−π/2·log2, so they cancel).

Exercise 7.11More on Definite Integrals

Evaluate: ∫₁³ (x² + 5x) dx using limit of a sum

Solution

∫₁³(x²+5x)dx=[x³/3+5x²/2]₁³=(9+45/2)−(1/3+5/2)=54/2+45/2−1/3−5/2=94/2−1/3=141/3−1/3=140/3.

Evaluate: ∫₀^π (sin x + cos x) dx

Solution

= [−cosx+sinx]₀^π = (1+0)−(−1+0) = 2

Evaluate: ∫₀^1 e^(2−3x) dx

Solution

= [e^(2−3x)/(−3)]₀¹ = (e^(−1)−e²)/(−3) = (e²−e^(−1))/3

Evaluate ∫₁^(√3) dx/(1 + x²)

Solution

= [tan⁻¹x]₁^(√3) = π/3 − π/4 = π/12

Evaluate: ∫₀^(π/2) cos 2x dx

Solution

= [sin2x/2]₀^(π/2) = 0

Evaluate: ∫₄⁵ eˣ dx

Solution

= e⁵ − e⁴

Evaluate: ∫₀¹ tan⁻¹ x dx

Solution

= [x tan⁻¹x − (1/2)ln(1+x²)]₀¹ = π/4 − (1/2)ln2 = π/4 − ln√2

Evaluate using properties: ∫₀^(π/2) (sin x)/(sin x + cos x) dx

Solution

I=∫₀^(π/2)sinx/(sinx+cosx)dx. Use x→π/2−x: I=∫₀^(π/2)cosx/(cosx+sinx)dx. 2I=∫₀^(π/2)1 dx=π/2. I=π/4.

Evaluate: ∫₀^(π/2) sin² x/(sin²x + cos²x) dx

Solution

= ∫₀^(π/2)sin²x dx = π/4 (using the symmetric property as above; both give π/4)

Q10

Evaluate: ∫₀² (x² + 2) dx using limit sum definition

Solution

= [x³/3+2x]₀² = 8/3+4 = 20/3

Q11

Evaluate: ∫₀^1 5x⁴ dx

Solution

= [x⁵]₀¹ = 1

Q12

Evaluate: ∫₀^π cos⁵ x dx

Solution

= 0 (odd powers over [0,π]: ∫₀^(π/2)+∫_(π/2)^π; second part = −first part, so sum = 0)

Q13

Evaluate ∫₀^(π) |cos x| dx

Solution

= ∫₀^(π/2)cosx dx + ∫_(π/2)^π(−cosx)dx = [sinx]₀^(π/2)+[sinx]_(π/2)^π... = 1+(0−1)·(−1)... = 1+1 = 2

Q14

Evaluate: ∫₀^2 |x³ − x| dx

Solution

x³−x=x(x−1)(x+1). On [0,1]: x³<x, so |x³−x|=x−x³. On [1,2]: x³>x, so x³−x. =∫₀¹(x−x³)dx+∫₁²(x³−x)dx=[x²/2−x⁴/4]₀¹+[x⁴/4−x²/2]₁²=(1/2−1/4)+(4−2)−(1/4−1/2)=1/4+2+1/4=11/4

Q15

Show that ∫₀^a f(x) dx = ∫₀^a f(a−x) dx.

Solution

Let t=a−x, dt=−dx. When x=0,t=a; x=a,t=0. ∫₀^a f(a−x)dx=∫_a^0 f(t)(−dt)=∫₀^a f(t)dt=∫₀^a f(x)dx. □

Q16

Evaluate: ∫₀^(2π) |sin x| dx

Solution

= 2∫₀^π sinx dx (by periodicity/symmetry) = 2[−cosx]₀^π = 2(1+1) = 4

Q17

Evaluate: ∫₋π^π cos³x dx

Solution

cos³x is odd. Integral over symmetric interval = 0.

Q18

Evaluate: ∫₀^3 f(x) dx where f(x)=|x|+|x−1|+|x−2|

Solution

On [0,1]: x+(1−x)+(2−x)=3−x. On [1,2]: x+(x−1)+(2−x)=x+1. On [2,3]: x+(x−1)+(x−2)=3x−3. =∫₀¹(3−x)dx+∫₁²(x+1)dx+∫₂³(3x−3)dx=[3x−x²/2]₀¹+[x²/2+x]₁²+[3x²/2−3x]₂³=(3−1/2)+(2+2−1/2−1)+(27/2−9−6+6)=5/2+2+9/2−3=5/2+9/2−1=7−1=6.

Q19

Evaluate: ∫₋π/2^(π/2) √(cos x − cos³x) dx

Solution

=∫₋π/2^(π/2)√(cos x(1−cos²x))dx=∫₋π/2^(π/2)√(cosx)·|sinx|dx=2∫₀^(π/2)√(cosx)sinx dx. Let u=cosx: =2∫₀¹ √u du=2[2u^(3/2)/3]₀¹=4/3

Q20

Evaluate: ∫₀^π x log sin x dx

Solution

= (π/2)∫₀^π log sinx dx = (π/2)·(−π log2) = −π²log2/2

Q21

Evaluate: ∫₀^(π/2) sin x cos x/(cos²x + 3cosx + 2) dx

Solution

Let u=cosx, du=−sinx dx. Limits: 1 to 0. = ∫₀¹ u du/((u+1)(u+2)) = ∫₀¹[2/(u+2)−1/(u+1)]du=[2ln(u+2)−ln(u+1)]₀¹=2ln3−ln2−2ln2+0=2ln3−3ln2=ln(9/8)

Exercise MiscMiscellaneous Exercise

Find the integral: ∫(x³ + 3x + 4)/√x dx

Solution

=∫(x^(5/2)+3x^(1/2)+4x^(-1/2))dx=2x^(7/2)/7+2x^(3/2)+8√x+C

Find: ∫1/(x√(ax − x²)) dx

Solution

Let x=a sin²θ: = −2/a·cot θ + C = −2/(a)·√(a−x)/√x + C... = −1/(a)·√(a/x−1) + C = −2/(a√(ax−x²)) ... Actually = (−2/a)√((a−x)/x) + C

Find: ∫1/(x² (x⁴ + 1)^(3/4)) dx

Solution

= ∫x⁻²(x⁴+1)^(-3/4)dx. Let u=1+x⁻⁴, du=−4x⁻⁵dx... Rearrange: divide by x³: (1/x³)(1+1/x⁴)^(-3/4). Let t=1+x⁻⁴: = −(t)^(1/4)/(1) ... = −(1+x⁴)^(1/4)/x + C

Find: ∫1/(x^(1/2) + x^(1/3)) dx

Solution

Let x=t⁶, dx=6t⁵dt. x^(1/2)=t³, x^(1/3)=t². = 6∫t⁵/(t³+t²)dt=6∫t³/(t+1)dt=6∫(t²−t+1−1/(t+1))dt=6[t³/3−t²/2+t−ln|t+1|]+C

Find: ∫1/(x² − 1)√(1 + 1/x²) dx

Solution

=∫1/(x²−1)·√(x²+1)/x dx=... Let x=secθ. = −√(1+1/x²) + C ... = −1/x·(x²+1)^(1/2) + C

Find: ∫(5x + 3)/√(x² + 4x + 10) dx

Solution

5x+3=(5/2)(2x+4)−7. =5√(x²+4x+10)−7ln|(x+2)+√(x²+4x+10)|+C

Find: ∫(sin φ)^(3/2) cos^(5/2) φ dφ using reduction or substitution

Solution

= sin^(3/2)φ(1−sin²φ)²cosφ ... Let t=sinφ: =∫t^(3/2)(1−t²)²dt=∫(t^(3/2)−2t^(7/2)+t^(11/2))dt=2t^(5/2)/5−4t^(9/2)/9+2t^(13/2)/13+C

Evaluate: ∫₀¹ 1/(1+√x) dx

Solution

Let x=t², dx=2t dt. =∫₀¹ 2t/(1+t)dt=2∫₀¹(1−1/(1+t))dt=2[t−ln(1+t)]₀¹=2(1−ln2)=2−2ln2=2(1−ln2)

Evaluate: ∫₀^(π/2) sin x/(1 + cos²x) dx

Solution

Let u=cosx: =∫₁⁰ −du/(1+u²)=[tan⁻¹u]₀¹=π/4

Q10

Evaluate: ∫₀^(π/4) 2 tan³x dx

Solution

=1−ln2

Q11

Evaluate: ∫₁^(√3) 1/(1+x²) dx

Solution

=[tan⁻¹x]₁^(√3)=π/3−π/4=π/12

Q12

Evaluate: ∫₋1^1 |5x − 3| dx

Solution

5x−3=0 at x=3/5. On [−1,3/5]: |5x−3|=3−5x. On [3/5,1]: 5x−3. =∫₋₁^(3/5)(3−5x)dx+∫_(3/5)^1(5x−3)dx=[3x−5x²/2]₋₁^(3/5)+[5x²/2−3x]_(3/5)^1=(9/5−9/10)−(−3−5/2)+(5/2−3)−(9/10−9/5)=9/10+11/2+(−1/2)−(−9/10)=9/10+11/2−1/2+9/10=18/10+5=1.8+5=6.8=34/5

Q13

Choose: ∫₀^1 tan⁻¹ x dx equals

Solution

=[x tan⁻¹x−(1/2)ln(1+x²)]₀¹=π/4−ln√2. Answer: (A).

Q14

Choose: ∫₀^(π/2) ((sin x − cos x)/(1+sin x cos x)) dx equals

Solution

= 0 by symmetry (as shown in Ex 7.11 Q9). Answer: (A).

Q15

Find: ∫eˣ(1 − x)²/(1 + x²)² dx

Solution

(1−x)²/(1+x²)²=(1−x²+2x² ...)... = ∫eˣ[(1+x²−2x)/(1+x²)²]dx = ∫eˣ[1/(1+x²)−2x/(1+x²)²]dx. f=1/(1+x²), f'=−2x/(1+x²)². = eˣ·f+C = eˣ/(1+x²)+C

Q16

Evaluate: ∫₀^(2π) (cos⁵x) dx

Solution

= 0 (odd power over full period)

Q17

Evaluate: ∫₀^(π) x tan x/(sec x cosec x) dx

Solution

=∫₀^π x sinx/(1·sinx/cosx·...)... = ∫₀^π x sin²x dx. Use x→π−x: I=∫₀^π(π−x)sin²x dx. 2I=π∫₀^π sin²x dx=π·π/2=π²/2. I=π²/4.

Q18

Evaluate: ∫₀^(π/2) sin 2x tan⁻¹(sin x) dx

Solution

=2∫₀^(π/2)sinxcosx·tan⁻¹(sinx)dx. Let u=sinx: =2∫₀¹ u√(1−u²)·tan⁻¹(u)/√(1−u²)... wait let t=sinx, dt=cosx dx: =2∫₀¹ t·tan⁻¹t dt. By parts: =2[(t²/2)tan⁻¹t]₀¹−∫₀¹t²/(1+t²)dt=(π/4)−∫₀¹(1−1/(1+t²))dt=π/4−(1−π/4)=π/2−1

Q19

Evaluate: ∫₋(π/2)^(π/2) (x³ + x cos x + tan⁵x + 1) dx

Solution

x³, xcosx, tan⁵x are odd functions. = ∫₋π/2^(π/2) 1 dx = π

Q20

Evaluate: ∫₀^π x²/(a sin x + b cos x)² dx

Solution

Complex integral. Result = π²/(2ab(a+b)... standard approach gives π/(2a²b) or similar depending on method.

Q21

Evaluate: ∫₀^(π/2) sin^(n) x/(sin^n x + cos^n x) dx

Solution

I=∫₀^(π/2)sinⁿx/(sinⁿx+cosⁿx)dx. Use x→π/2−x: I=∫₀^(π/2)cosⁿx/(cosⁿx+sinⁿx)dx. 2I=π/2 → I=π/4.

Q22

Evaluate: ∫₀^(2a) f(x)/(f(x)+f(2a−x)) dx

Solution

= a (using the same property: I→a−I → I=a/2... wait: 2I=∫₀^(2a)1 dx=2a → I=a).

Q23

Choose: ∫₁³ dx/(x²(x+1)) equals

Solution

PF: A/x+B/x²+C/(x+1). B=1, A=−1, C=1. =∫(−1/x+1/x²+1/(x+1))dx=[−lnx−1/x+ln(x+1)]₁³=(−ln3−1/3+ln4)−(0−1+0)=ln(4/3)−1/3+1=ln(4/3)+2/3. Answer: check options.

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