# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.9 | Set 1

• Last Updated : 30 Apr, 2021

### Question 1. ∫(log⁡x)/x dx

Solution:

Given that, I = ∫(log⁡x)/x dx

Let us considered log⁡x = t

Now differentiating both side we get,

d(log⁡x) = dt

1/x dx = dt

dx = xdt

Then, put log⁡x = t and dx = xdt, we get

I = ∫ t/x × (x)dt

= ∫ tdt

= t2/2 + c

= (log⁡x)2/2 + c

Hence, I = (log⁡x)2/2 + c

### Question 2.∫(log⁡(1 + 1/x))/(x(1 + x)) dx

Solution:

Given that I = ∫(log⁡(1 + 1/x))/(x(1 + x)) dx   ……(i)

Let us considered log⁡(1 + 1/x) = t then

On differentiating both side we get,

d[log⁡(1 + 1/x)]=dt

1/(1 + 1/x) × (-1)/x2dx = dt

1/((x + 1)/x) × (-1)/x2dx = dt

(-x)/(x2 (x + 1)) dx = -dt

dx/(x(x + 1)) = -dt

Now, on putting log⁡(1 + 1/x) = t and dx/(x(x + 1)) = -dt in equation (i), we get

I = ∫t × -dt

= -t2/2 + c

= -1/2 [log⁡(1 + 1/x)]2 + c

Hence, I = -1/2 [log⁡(1 + 1/x)]2 + c

### Question 3. ∫((1 + √x )2)/√x dx

Solution:

Given that I = ∫((1 + √x)2)/√x dx

Let us considered (1 + √x) = t then,

On differentiating both side we get,

d(1 + √x) = dt

1/(2√x) dx = dt

dx = dt × 2√x

Now on putting (1 + √x) = t and dx = dt × 2√x, we get

I = ∫t2/√x × dt × 2√x

= 2∫t2dt

= 2 × t3/3 + c

= 2/3[1 + √x]3 + c

Hence, I = 2/3(1 + √x)3 + c

### Question 4. ∫√(1 + ex) ex d

Solution:

Given that I = ∫√(1 + ex) ex dx  ……(i)

Let us considered 1 + ex = t then,

On differentiating both side we get,

d(1 + ex) = dt

ex dx = dt

dx = dt/ex

Now on putting 1 + ex = t and dx = dt/ex in equation (i), we get

I = ∫√t × ex × dt/ex

= ∫ t1/2 dt

= 2/3t3/2 + c

= 2/3 (1 + ex)3/2+c

### Question 5. ∫∛(cos2x) sin⁡x dx

Solution:

Given that I = ∫∛(cos2x) sin⁡x dx    ……(i)

Let us considered cos⁡x = t then,

On differentiating both side we get,

d(cos⁡x) = dt

-sin⁡xdx = dt

dx = -dt/(sin⁡x))

Now on putting cos⁡x = t and dx = -dt/(sin⁡x) in equation (i), we get

I = ∫∛(t2) sin⁡x × (-dt)/(sin⁡x)

= -∫t2/3 sin⁡x dt/(sin⁡x)

= -∫ t2/3 dt

= -3/5 × t5/3

Hence, I = -3/5(cos⁡x)5/3 + c

### Question 6. ∫ex/(1 + ex)2 dx

Solution:

Given that I = ∫ex/(1 + ex)2 dx   …….(i)

Let us considered 1 + ex = t then,

On differentiating both side we get,

d(1 + ex) = dt

ex dx = dt

dx = dt/ex

Now on putting 1 + ex = t and dx = dt/ex in equation (i), we get

I = ∫ex/t2 × dt/ex

= ∫dt/t2

= ∫t-2 dt

= -t-1 + c

= -1/t + c

= -1/(1 + ex) + c

Hence, I = -1/(1 + ex) + c

### Question 7. ∫cot3⁡x cosec2⁡x dx

Solution:

Given that I = ∫cot3⁡x cosec2⁡x dx   …….(i)

Let us considered cot⁡x = t then,

On differentiating both side we get,

d(cotx) = dt

-cosec2x dx = dt

dx = -dt/cosec2x

Now on putting cot⁡x = t and dx = -dt/(cosec2⁡x) in equation (i), we get

I = ∫ t3 cosec2x × (-dt)/(cosec2⁡x)

= -∫ t3 dt

= -t4/4 + c

= -(cot4⁡x)/4 + c

Hence, I = -(cot4⁡x)/4 + c

### Question 8.

Solution:

Given that I =[Tex] [/Tex] …….(i)

Let us considered sin-1⁡x = t then,

On differentiating both side we get,

d(sin-1x) = dt

1/√(1 – x2)dx = dt

dx = √(1 – x2) dt)

Now on putting sin-1x = t and dx = √(1 – x2) dt in equation (i), we get

I = ∫ (et)2/√(1 – x2) × √(1 – x2) dt

= ∫e2t dt

= e2t/2 + c

+ c

Hence, I =  + c

### Question 9. ∫(1 + sin⁡x)/√(x – cos⁡x) dx

Solution:

Given that I = ∫(1 + sin⁡x)/√(x – cos⁡x) dx    ……..(i)

Let us considered x – cos⁡x = t, then

On differentiating both side we get,

d(x – cos⁡x) = dt

[1 – (-sin⁡x)]dx = dt

(1 + sin⁡x)dx = dt

Now on putting x – cos⁡x = t and (1 + sin⁡x)dx = dt in equation (i), we get

I = ∫ dt/√t

= ∫ t-1/2 dt

= 2t1/2 + c

= 2(x – cos⁡x)1/2 + c

Hence, I = 2√(x – cos⁡x) + c

### Question 10.  ∫1/(√(1 – x2) (sin-1x)2) dx

Solution:

Given that I = ∫1/(√(1 – x2) (sin-1⁡x)2) dx   …..(i)

Let us considered sin-1⁡x = t then,

On differentiating both side we get,

d(sin-1⁡x) = dt

1/√(1 – x2 ) dx = dt

Now on putting sin-1⁡x = t and 1/√(1 – x2) dx = dt in equation (i), we get

I = ∫dt/t2

= ∫t-2 dt

= -t-1 + c

= (-1)/t + c

= (-1)/(sin-1⁡x) + c

Hence, I = (-1)/(sin-1⁡x) + c

### Question 11. ∫(cot⁡x)/√(sin⁡x) dx

Solution:

Given that I = ∫(cot⁡x)/√(sin⁡x) dx    …….(i)

Let us considered sin⁡x = t then,

On differentiating both side we get,

d(sin⁡x) = dt

cos⁡xdx = dt

Now, I = ∫(cot⁡x)/√(sin⁡x) dx

= ∫(cos⁡x)/(sin⁡x√(sin⁡x)) dx

= ∫ cos⁡x/(sin⁡x)3/2dx

= ∫ cos⁡x/(sin⁡x)3/2 dx   …….(ii)

Now on putting sin⁡x = t and cos⁡xdx = dt in equation (ii), we get

I = ∫ dt/t3/2

= ∫ t-3/2 dt

= -2t-1/2 + c

= -2/√t + c

= -2/√(sin⁡x) + c

I = -2/√sin⁡x + c

### Question 12. ∫(tan⁡x)/√(cos⁡x) dx

Solution:

Given that I = ∫(tan⁡x)/√(cos⁡x) dx

I = ∫sin⁡x/cos⁡x√(cos⁡x) dx

= ∫ sin⁡x/(cos⁡x)3/2 dx

= ∫sin⁡x/(cos⁡x)3/2 dx   ……..(i)

Let us considered cos⁡x = t then,

On differentiating both side we get,

d(cos⁡x) = dt

-sin⁡xdx = dt

sin⁡xdx = -dt

Now on putting cos⁡x = t and sin⁡xdx = -dt in equation (i), we get

I = ∫(-dt)/t-3/2

= -∫t-3/2 dt

= -[-2t-1/2] + c

= 2/t1/2 + c

= 2/√(cos⁡x) + c)

Hence, I = 2/√(cos⁡x) + c

### Question 13. ∫cos3x/√(sin⁡x) dx

Solution:

Given that I = ∫cos3⁡x/√(sin⁡x) dx

= ∫(cos2xcos⁡x)/√(sin⁡x) dx

= ∫((1 – sin2⁡x)cos⁡x)/√(sin⁡x) dx

= ∫((1 – sin2x))/√(sin⁡x) cos⁡xdx ……(i)

Let us considered  sin⁡x = t then,

On differentiating both side we get,

d(sin⁡x) = dt

cos⁡xdx = dt

Now on putting sin⁡x = t and cos⁡xdx = dt in equation (i), we get

I = ∫(1 – t2)/√t dt

= ∫(t-1/2-t2 x t-1/2)dt

=∫(t-1/2 – t3/2)dt

= 2t1/2 – 2/5 t5/2 + c

= 2(sin⁡x)1/2 – 2/5(sin⁡x)5/2 + c

Hence, I = 2√(sin⁡x) – 2/5(sin⁡x)5/2 + c

### Question 14. ∫(sin3x)/√(cos⁡x) dx

Solution:

Given that I = ∫(sin3⁡x)/√(cos⁡x) dx

= ∫(sin2⁡xsin⁡x)/√(cos⁡x) dx

=∫((1 – cos2⁡x))/√(cos⁡x) sin⁡xdx …….(i)

Let us considered cos⁡x = t then,

On differentiating both side we get,

d(cos⁡x) = dt

-sin⁡xdx = dt

sin⁡xdx = -dt

Now on putting cos⁡x = t and sin⁡xdx = -dt in equation (i), we get

I = ∫((1 – t2))/√t × -dt

= ∫(t2 – 1)/√t dt

= ∫(t2/t1/2 – 1/t1/2)dx

= ∫(t2-1/2 – t-1/2)dt

= ∫(t3/2 – t1/2)dt

= 2/5 t5/2 – 2t1/2 + c

= 2/5 cos5/2⁡x – 2cos1/2⁡x + c

Hence, I = 2/5 cos5/2x – 2√(cos⁡x) + c

### Question 15. ∫1/(√(tan-1x) (1 + x2)) dx

Solution:

Given that I = ∫1/(√(tan-1x) (1 + x2)) dx  …..(i)

Let us considered tan-1⁡x = t, then

On differentiating both side we get,

d(tan-1⁡x) = dt

1/(1 + x2) dx = dt

Now on putting tan-1⁡x = t and 1/(1 + x2) dx = dt in equation (i), we get

I = ∫1/√t dt

= ∫t-1/2 dt

= 2t1/2 + c

= 2√tan-1⁡x + c

Hence, I = 2√tan-1⁡x + c

### Question 16. ∫√(tan⁡x)/(sin⁡xcos⁡x) dx

Solution:

Given that I = ∫√(tan⁡x)/(sin⁡xcos⁡x) dx

= ∫(√(tan⁡x)×cos⁡x)/(sin⁡xcos⁡x×cos⁡x) dx

= ∫√(tan⁡x)/(tan⁡xcos2⁡x) dx

= ∫(sec2⁡xdx)/√(tan⁡x) dx

Let us considered tan⁡x = t, then

On differentiating both side we get,

sec2⁡xdx = dt

Now

I = ∫ dt/√t

= 2√t + c

Hence, I = 2√tan⁡x + c

### Question 17. 1/x × (log⁡x)2 dx

Solution:

Given that I = ∫1/x × (log⁡x)2 dx   …..(i)

Let us considered log⁡x = t then,

On differentiating both side we get,

d(log⁡x) = dt

1/x dx = dt

Now on putting log⁡x = t and 1/x dx = dt in equation (i), we get

I = ∫t2 dt

= t3/3 + c

= (log⁡x)3/3 + c

Hence, I = (log⁡x)3/3 + c

### Question 18. ∫sin5x cos⁡x dx

Solution:

Given that I = ∫sin5⁡x cos⁡x dx  ……(i)

Let us considered sin⁡x = t then,

On differentiating both side we get,

d(sin⁡x) = dt

cos⁡xdx = dt

Now on putting sin⁡x = t and cos⁡xdx = dt in equation (i), we get

I = ∫ t5 dt

= t6/6 + c

= (sin6⁡x)/6 + c

Hencec, I = 1/6 (sin6⁡x) + c

### Question 19. ∫tan3/2⁡x sec2⁡x dx

Solution:

Given that I = ∫tan3/2⁡xsec2⁡xdx  ……(i)

Let us considered tan⁡x = t then,

On differentiating both side we get,

d(tan⁡x) = dt

sec2⁡xdx = dt

Now on putting tan⁡x = t and sec2xdx = dt in equation (i), we get

I = ∫ t3/2 dt

= 2/5 t5/2 + c

= 2/5(tan⁡x)5/2 + c

Hence, I = 2/5 tan5/2⁡x + c

### Question 20. ∫(x3)/(x2 + 1)2dx

Solution:

Given that I = ∫(x3)/(x2 + 1)2dx …….(i)

Let us considered 1 + x2 = t then,

On differentiating both side we get,

d(1 + x2) = dt

2xdx = dt

xdx = dt/2

Now on putting 1 + x2 = t and xdx = dt/2 in equation (i),we get

I = ∫x2/t3 × dt/2

= 1/2∫(t – 1)/t3 dt         [1 + x2 = t]

= 1/2∫[(t/t3 – 1/t3)dt]

= 1/2∫(t-2 – t-3)dt

= 1/2 [-1t-1 – t-2/(-2)] + c

= 1/2 [-1/t + 1/(2t2)] + c

= -1/2t + 1/(4t2) + c

= -1/2(1 + x2) + 1/(4(1 + x2)2) + c

= (-2(1 + x2) + 1)/(4(1 + x2)2) + c

= (-2 – 2x2 + 1)/(4(1 + x2)2) + c

= (-2x2 – 1)/(4(1 + x2)2) + c

= -(1 + 2x2 )/(4(x2 + 1)2) + c

Henec, I = -(1 + 2x2)/(4(x2 + 1)2) + c

### Question 21. ∫(4x + 2)√(x2 + x + 1) dx

Solution:

Given that I = ∫(4x + 2)√(x2 + x + 1) dx

Let us considered x2 + x + 1 = t then,

On differentiating both side we get,

(2x + 1)dx = dt

Now,

I = ∫ (4x + 2)√(x2 + x + 1) dx

= ∫2√t dt

= 2∫√t dt

= 2t3/2/(3/2) + c

Hence, I = 4/3 (x2 + x + 1)3/2 + c

### Question 22. ∫(4x + 3)/√(2x2 + 3x + 1) dx

Solution:

Given that l = ∫(4x + 3)/√(2x2 + 3x + 1) dx  ……(i)

Let us considered 2x2 + 3x + 1 = t then,

On differentiating both side we get,

d(2x2 + 3x + 1) = dt

(4x + 3)dx = dt

Now on putting 2x2 + 3x + 1 = t and (4x + 3)dx = dt in equation (i), we get

I = ∫dt/√t

= ∫t-1/2 dt

= 2t1/2 + c

= 2√t + c

Hence, I = 2√(2x2 + 3x + 1) + c

### Question 23. ∫1/(1 + √x) dx

Solution:

Given that I = ∫1/(1 + √x) dx   …….(i)

Let us considered x = t2 then,

On differentiating both side we get,

dx = d(t2)

dx = 2tdt

Now on putting x = t2 and dx = 2tdt in equation (i), we get

I = ∫2t/(1 + √(t2)) dt

= ∫2t/(1 + t) dt

= 2∫t/(1 + t) dt

= 2∫(1 + t – 1)/(1 + t) dt

= 2⌋[(1 + t)/(1 + t) – 1/(1 + t)]dt

= 2∫dt – 2∫1/(1 + t) dt

= 2t – 2log⁡|(1 + t)| + c

= 2√x – 2log⁡|(1 + √x)| + c

Hence, I = 2√x – 2log⁡|(1 + √x)| + c

### Question 24.

Solution:

Given that I =  …….(i)

Let us considered cos2x = t then,

On differentiating both side we get,

d(cos2⁡x) = dt

-2cos⁡x sin⁡x dx = dt

-sin⁡2x dx = dt

sin⁡2x dx = -dt

Now on putting cos2⁡x = t and sin⁡2x dx = -dt in equation (i), we get

I = ∫et(-dt)

= -et + c

= – + c

Hence, I = – + c

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