# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.2 | Set 2

### Question 25. Evaluate ∫(tanx + cotx)^{2} dx

**Solution:**

We have,∫(tanx + cotx)^{2}dxBy using formula (x + y)

^{2}= x^{2}+ y^{2}+ 2xyWe get, ∫(tan

^{2}x + cot^{2}x + 2tanx cotx)dx= ∫ (sec

^{2}x – 1 + cosec^{2}x – 1 + ((2 × 1)/cotx) × cotx)dx= ∫ (sec

^{2}x + cosec^{2}x)dx= ∫sec

^{2}xdx + ∫cosec^{2}xdx= tanx – cotx + c

### Question 26. Evaluate ∫(1 – cos2x)/(1 + cos2x) dx

**Solution:**

We have,∫(1 – cos2x)/(1 + cos2x) dx= ∫(2sin

^{2}x)/(2cos^{2}x) dx= ∫tan

^{2}xdx= ∫(sec

^{2}x – 1)dx= ∫sec

^{2}xdx – 1∫dx= tanx – x + c

### Question 27. Evaluate ∫(cosx)/(1 – cosx) dx

**Solution:**

We have,∫(cosx)/(1 – cosx) dx= ∫(cosx(1 + cosx))/((1 – cosx)(1 + cosx)) dx

= ∫(cosx + cos

^{2}x)/(1 – cos^{2}x) dx= ∫(cosx + cos

^{2}x)/(sin^{2}x) dx= ∫(cosx)/(sin

^{2}x) dx + ∫(cos^{2}x)/(sin^{2}x) dx [Since, cosx/sinx = cotx]= ∫cotx × cosecxdx + ∫(cosec

^{2}x – 1)dx [Since, cot^{2}x = cosec^{2}x – 1]= -cosecx – cotx – x + c

### Question 28. Evaluate ∫cos^{2}x – sin^{2}x/√(1 + cos4x) dx

**Solution:**

We have,∫cos^{2}x – sin^{2}x/√(1 + cos4x) dx= ∫(cos

^{2}x – sin^{2}x)/√(2cos^{2}2x) dx= 1/√2 ∫(cos

^{2}x – sin^{2}x)/(cos2x) dx= 1/√2∣(cos

^{2}x – sin^{2}x)/(cos^{2}x – sin^{2}x) dx= 1/√2∫1 × dx

= x/√2 + c

### Question 29. Evaluate ∫ 1/(1 – cosx) dx

**Solution:**

We have, ∫ 1/(1 – cosx) dx

= ∫1/(1 – cosx) × (1 + cosx)/(1 + cosx) × dx

= ∫(1 + cosx)/(1 – cos

^{2}x) × dx= ∫(1 + cosx)/(sin

^{2}x) × dx= ∫1/(sin

^{2}x) dx + ∫(cosx)/(sin^{2}2x) dx= ∫cosec

^{2}xdx + ∫cotx × cosecx dx= -cotx – cosecx + c

### Question 30. Evaluate ∫1/(1 – sinx) dx

**Solution:**

We have, ∫1/(1 – sinx) dx

= ∫1/(1 – sinx) × (1 + sinx)/(1 + sinx) × dx

= ∫(1 + sinx)/(1 – sin

^{2}x) × dx= ∫(1 + sinx)/(cos

^{2}x) × dx= ∫(1/(cos

^{2}x) + (sinx)/(cos^{2}x)) × dx= ∫1/(cos

^{2}x) dx + ∫(sinx)/(cos^{2}x) × dx= ∫sec

^{2}xdx + ∫tanx secx dx= tanx + secx + c

### Question 31. Evaluate ∫(tanx)/(secx + tanx) dx

**Solution:**

We have, ∫(tanx)/(secx + tanx) dx

= ∫(tanx)/(secx + tanx) × (secx – tanx)/(secx – tanx) × dx

= ∫(tanx(secx – tanx))/(sec

^{2}x – tan^{2}x) × dx= ∫(tanxsecx – tan

^{2}x)dx= ∫sectanxdx – ∫(sec

^{2}x – 1)dx= ∫secxtanxdx – ∫sec

^{2}xdx + 1∫dx= secx – tanx + x + c

### Question 32. Evaluate ∫(cosecx)/(cosecx – cotx)dx

**Solution:**

We have, ∫(cosecx)/(cosecx – cotx)dx

= ∫(cosecx)/(cosecx – cotx) × (cosecx + cotx)/(cosecx + cotx) × dx

= ∫(cosecx(cosecx + cotx))/(cosec

^{2}x – cot^{2}x) × dx= ∫(cosec

^{2}x + cosecx cotx)dx= ∫cosec

^{2}xdx + ∫cosecx cotx dx= -cotx – cosecx + c

### Question 33. Evaluate ∫1/(1 + cos2x) dx

**Solution:**

We have, ∫1/(1 + cos2x) dx

= ∫ 1/(2cos

^{2}x) × dx= 1/2 ∫sec

^{2}x × dx= 1/2 × tanx + c

= (tanx)/2 + c

### Question 34. Evaluate∫1/(1 – cos2x) dx

**Solution:**

We have, ∫1/(1 – cos2x) dx

= ∫1/(2sin

^{2}x)dx= 1/2 ∫cosec

^{2}x dx= (-1)/2 × cotx + c

= (-cotx)/2 + c

### Question 35. Evaluate ∫tan^{-1}[(sin2x)/(1 + cos2x)]dx

**Solution:**

We have, ∫tan

^{-1}[(sin2x)/(1 + cos2x)]dx= ∫tan

^{-1}[(2sinxcosx)/(2cos^{2}x)]dx= ∫tan

^{-1}[(sinx)/(cosx)]dx= ∫tan

^{-1}(tanx)dx= ∫xdx

= x

^{2}/2 + c

### Question 36. Evaluate ∫cos^{-1}(sinx)dx

**Solution:**

We have, ∫cos

^{-1}(sinx)dx= ∫cos

^{-1}[cos(π/2 – x)]dx= ∫(π/2 – x)dx

= π/2 ∫dx – ∫xdx

= π/2 × x – x

^{2}/2 + c

### Question 37. Evaluate ∫ cot^{-1}(sinx)dx

**Solution:**

We have, ∫ cot

^{-1}(sinx)dx= ∫cot

^{-1}[(sin2x)/(1 – cos2x)]dx= ∫cot

^{-1}((cosx)/(sinx))dx= ∫cot

^{-1}(cotx)dx= ∫xdx

= x

^{2}/2 + c

### Question 38. Evaluate ∫ sin^{-1}((2tanx)/(1 + tan^{2}x))dx

**Solution:**

We have, ∫ sin

^{-1}((2tanx)/(1 + tan^{2}x))dx= ∫ sin

^{-1}(sin2x)dx= ∫2xdx

= 2∫xdx

= (2x

^{2})/2 + c= x

^{2 }+ c

### Question 39. Evaluate ∫((x^{3 }+ 8)(x – 1))/(x^{2 }– 2x + 4) dx

**Solution:**

We have, ∫((x

^{3 }+ 8)(x – 1))/(x^{2 }– 2x + 4) dx= ∫((x + 2)(x

^{2 }– 2x + 4)(x – 1))/(x^{2 }– 2x + 4) dx= ∫(x + 2)(x – 1)dx

= ∫(x

^{2 }– x+2x – 2)dx= ∫(x

^{2 }+ x – 2)dx= x

^{3}/3 + x^{2}/2 – 2x + c

### Question 40. Evaluate ∫(atanx + bcotx)^{2} dx

**Solution:**

We have, ∫(atanx + bcotx)

^{2}dxBy using formula (x + y)

^{2}= x^{2}+ y^{2}+ 2xy , we get= ∫(a

^{2}tan^{2}x + b^{2}cot^{2}x + 2ab tanx cotx)dx= ∫[a

^{2}(sec^{2}x – 1) + b^{2}(cosec^{2}x – 1) + 2ab]dx= ∫[a

^{2}sec^{2}x – a^{2 }+ b^{2}cosec^{2}x – b^{2 }+ 2ab]dx= a

^{2}tanx – a^{2}x – b^{2}cotx – b^{2}x + 2abx + c= a

^{2}tanx – b^{2}cotx – (a^{2 }+ b^{2 }– 2ab)x + c

### Question 41. Evaluate ∫(x^{3 }– 3x^{2 }+ 5x – 7 + x^{2} a^{x})/(2x^{2}) dx

**Solution:**

We have, ∫(x

^{3 }– 3x^{2 }+ 5x – 7 + x^{2}a^{x})/(2x^{2}) dx= 1/2 ∫x

^{3}/x^{2}dx – 3/2∫x^{2}/x^{2}dx + 5/2∫x/x^{2}dx – 7/2∫x^{-2}dx + 1/2∫(x^{2}a^{x})/x^{2}dx= 1/2 × x

^{2}/2 – 3/2x + 5/2 logx – 7/2 x^{-1 }+ 1/2a^{x}/(loga) + c= 1/2 [x

^{2}/2 – 3x + 5logx + 7/x + a^{x}/(loga)] + c

### Question 42. Evaluate ∫cosx/(1 + cosx) dx

**Solution:**

We have, ∫cosx/(1 + cosx) dx …..(1)

Now solve

Since, cosx = cos

^{2}x/2 – sin^{2}x/2 and cosx + 1 = 2cos^{2}x/2So, we get cosx/(1 + cosx) = 1/2[1 – tan

^{2}x/2]Now put this value in eq(1), we get

= 1/2 ∫(1 – tan

^{2}x/2)dx= 1/2 ∫(1 – sec

^{2}x/2 + 1)dx= 1/2 ∫(2 – sec

^{2}x/2)dx= 1/2 [2x – (tanx/2)/(1/2)] + c

= x – tanx/2 + c

### Question 43. Evaluate∫(1 – cosx)/(1 + cosx) dx

**Solution:**

We have, ∫(1 – cosx)/(1 + cosx) dx ….(1)

Now solve

(1 – cosx)/(1 + cosx) = (2sin

^{2}x)/(2cos^{2}x)= tan

^{2}x/2= (sec

^{2}x/2 – 1) [Since, 2sin^{2}x/2 = 1 – cosx and 2cos^{2}x/2 = 1 + cosx]Now put this value in eq(1), we get

= ∫(sec

^{2}x/2 – 1)dx= tan(x/2)/(1/2) – x + c

= 2tanx/2 – x + c

### Question 44. Evaluate ∫{3sinx – 4cosx + 5/(cos^{2}x) – 6/(sin^{2}x) + tan^{2}x – cot^{2}x}dx

**Solution:**

We have, ∫{3sinx – 4cosx + 5/(cos

^{2}x) – 6/(sin^{2}x) + tan^{2}x – cot^{2}x}dx= 3∫sinxdx – 4∫cosxdx + 5∫sec

^{2}dx – 6∫cosec^{2}x + ∫tan^{2}xdx – ∫cot^{2}xdx= 3∫sinxdx – 4∫cosxdx + 5∫sec

^{2}xdx – 6∫cosec^{2}x + ∫(sec^{2}x – 1)dx – ∫(cosec^{2}x – 1)dx= 3∫sinxdx – 4∫cosxdx + 6∫sec

^{2}xdx – 7∫cosec^{2}xdx= -3cosx – 4sinx + 6tanx + 7cotx + c

### Question 45. If f'(x) = x – 1/x^{2} and f(1) = 1/2, find f(x)?

**Solution:**

Given that ∫f'(x) = x – 1/x

^{2}and f(1) = 1/2

We have to find f(x)

So, ∫f'(x) = ∫xdx – ∫1/x

^{2}dxf(x) = x

^{2}/2 + x^{-1 }+ cf(x) = x

^{2}/2 + 1/x + cf(x) = x

^{2}/2 + 1/x + c …..(i)As we know that

f(1) = 1/2

1

^{2}/2 + 1/1 + c = 1/21/2 + 1 + c = 1/2

c = -1

On putting c = -1 in (i), we get

f(x) = x

^{2}/2 + 1/x – 1

### Question 46. If f'(x) = x + b, f(1) = 5, f(2) = 13, find f(x)?

**Solution:**

Given that f'(x) = x + b

and f(1) = 5, f(2) = 13

We have to find f(x)

So, ∫f'(x) = ∫(x + b)dx

f(x) = x

^{2}/2 + bx + c …….(i)As we know that

f(1) = 5

1

^{2}/2 + b × 1 + c = 51/2 + b + c = 5

b + c = 9/2 …….(ii)

Also, f(2) = 13

2

^{2}/2 + b × 2 + c = 132 + 2b + c = 13

2b + c = 11 …….(iii)

Now, subtract eq(ii) from eq(iii), we get

b = 11 – 9/2

b = 13/2

Now, put b = 13/2 in eq(ii), we get

13/2 + c = 9/2

c = 9/2 – 13/2

c = (9 – 13)/2

= (-4)/2

= -2

Now, on putting b = 13/2 and c = -2 in equation (i), we get

f(x) = x

^{2}/x + 13/2x – 2f(x) = x

^{2}/2 + 13/2x – 2

### Question 47. If f'(x) = 8x^{3 }– 2x, f(2) = 8, find f(x)?

**Solution:**

Given that f'(x) = 8x

^{3 }– 2xand f(2) = 8

We have to find f(x)

So, ∫f'(x)dx = ∫(8x

^{3 }– 2x)dxf(x) = ∫(8x

^{3 }– 2x)dx= ∫8x

^{3}dx – ∫2xdx= (8x

^{4})/4 – (2x^{2})/2 + c= 2x

^{4 }– x^{2 }+ cf(x) = 2x

^{4 }– x^{2 }+ c ……….(i)As we know that f(2) = 8

So, f(2) = 2(2)

^{4 }– (2)^{2 }+ c = 832 – 4 + c = 8

28 + c = 8

c = -20

Now, Put c = -20 in eq(i), we get

f(x) = 2x

^{4 }– x^{2 }– 20

### Question 48. If f'(x) = asinx + bcosx and f'(0) = 4, f(0) = 3, f(π/2) = 5, find f(x)?

**Solution:**

Given that, f'(x) = asinx + bcosx

and f'(0) = 4, f(0) = 3, f(π/2) = 5

We have to find f(x)

So,

∫f'(x) = ∫(asinx + bcosx)dx

f(x) = -acosx + bsinx + c

f(x) = -acosx + bsinx + c ………(i)

As we know that f'(0) = 4

So, f'(0) = asin0 + bcos0 = 4

a × 0 + b × 1 = 4

b = 4

Also, f(0) = 3

f(0) = -acos0 + bsin0 + c = 3

-a + 0 + c = 3

c – a = 3 ……..(ii)

Also, f(π/2) = 5

f(π/2) = -acos(π/2) + bsin(π/2) + c = 5

-a × 0 + b × 1 + c = 5

b + c = 5

4 + c = 5 [Since, b = 4]

c = 5 – 4

c = 1

Now, put c = 1 in eq(ii), we get 1 – a = 3

-a = 3 – 1

-a = 2

a = -2

Now, put a = -2, b = 4, and c = 1 in eq(i), we get

f(x) = -(-2)cosx + 4sinx + 1

f(x) = 2cosx + 4sinx + 1

### Question 49. Write the primitive or anti-derivative of f(x) = √x + 1/√x.

**Solution:**

We have, f(x) = √x + 1/√x

∫f(x) = ∫(√x + 1/√x)dx

= ∫x

^{1/2}dx + ∫ x^{-1/2}dx= 2/3 x

^{3/2 }+ 2x^{1/2 }+ cHence, the primitive or anti-derivative of f(x) is 2/3 x

^{3/2 }+ 2x^{1/2 }+ c.

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