# RD Sharma Class 12 Ex 19.9 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.9 Solutions Chapter 19 Indefinite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest Sharma Class 12 Ex 19.9 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 19.9 Solutions Chapter 19 Indefinite Integrals

### 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)]+ c

Hence, I = -1/2 [log⁡(1 + 1/x)]+ 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]+ c

Hence, I = 2/3(1 + √x)+ c

### Question 4. ∫√(1 + ex) ex d

Solution:

Given that I = ∫√(1 + ex) edx  ……(i)

Let us considered 1 + e= t then,

On differentiating both side we get,

d(1 + ex) = dt

ex dx = dt

dx = dt/ex

Now on putting 1 + e= 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)dx   …….(i)

Let us considered 1 + e= t then,

On differentiating both side we get,

d(1 + ex) = dt

ex dx = dt

dx = dt/ex

Now on putting 1 + e= t and dx = dt/ein 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 = ∫ tcosec2x × (-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

= ∫(t– 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)/(x+ 1)2dx …….(i)

Let us considered 1 + x= t then,

On differentiating both side we get,

d(1 + x2) = dt

2xdx = dt

xdx = dt/2

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

I = ∫x2/t3 × dt/2

= 1/2∫(t – 1)/t3 dt         [1 + x= 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 – 2x+ 1)/(4(1 + x2)2) + c

= (-2x– 1)/(4(1 + x2)2) + c

= -(1 + 2x2 )/(4(x+ 1)2) + c

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

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

Solution:

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

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

On differentiating both side we get,

(2x + 1)dx = dt

Now,

I = ∫ (4x + 2)√(x+ x + 1) dx

= ∫2√t dt

= 2∫√t dt

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

Hence, I = 4/3 (x+ x + 1)3/2 + c

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

Solution:

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

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

On differentiating both side we get,

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

(4x + 3)dx = dt

Now on putting 2x+ 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√(2x+ 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 = tand 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)

= -e+ c

= – + c

Hence, I = – + c

Question 25. ∫(1 + cos⁡x)/((x + sin⁡x)3) dx
Solution:

Given that I = ∫(1 + cos⁡x)/((x + sin⁡x)3) dx    …..(i)
Let us considered x + sin⁡x = t then,
On differentiating both side we get,
d(x + sin⁡x) = dt
(1 + cos⁡x)dx = dt
Now on putting x + sin⁡x = t and (1 + cos⁡x)dx = dt in equation (i), we get
I = ∫ dt/t3
= ∫ t-3 dt
= t-2/-2 + c
= -1/(2t2) + c
= (-1)/(2(x + sin⁡x)2) + c
Hence, I = (-1)/(2(x + sin⁡x)2) + c
Question 26. ∫(cos⁡x – sin⁡x)/(1 + sin⁡2x) dx
Solution:
Given that I = (cos⁡x – sin⁡x)/(1 + sin⁡2x)
= (cos⁡x – sin⁡x)/((sin2⁡x + cos2⁡x) + 2sin⁡xcos⁡x)  [Because sin2⁡x + cos2⁡x = 1 and sin⁡2x = 2sin⁡xcos⁡x]
Let us considered sin⁡x + cos⁡x = t
On differentiating both side we get,
(cos⁡x – sin⁡x)dx = dt
Now,
= ∫(cos⁡x – sin⁡x)/(1 + sin⁡2x) dx
= ∫(cos⁡x – sin⁡x)/((sin⁡x + cos⁡x)2) dx
= ∫dt/t2
= ∫t-2 dt
= -t-1 + c
= -1/t + c
Hence, I = (-1)/(sin⁡x + cos⁡x) + c
Question 27. ∫(sin⁡2x)/(a + bcos⁡2x)dx
Solution:
Given that I = ∫(sin⁡2x)/((a + bcos⁡2x)2) dx   ……(i)
Let us considered a + bcos⁡2x = t then,
On differentiating both side we get,
(a + bcos⁡2x) = dt
b(-2sin⁡2x)dx = dt
sin⁡2x dx = -dt/2b
Now on putting a + bcos⁡2x = t and sin⁡2xdx = -dt/2b in equation (i), we get
I = ∫1/t× (-dt)/2b
= (-1)/2b ∫ t-2 dt
= -1/2b (-1t-1) + c
= 1/2bt + c
= 1/(2b(a + bcos⁡2x)) + c
Hence, I = 1/(2b(a + bcos⁡2x)) + c
Question 28. ∫(log⁡x2)/x dx
Solution:
Given that I = ∫(log⁡x2)/x dx  ……..(i)
Let us considered log⁡x = t then,
On differentiating both side we get,
d(log⁡x) = dt
1/x dx = dt
dx/x = dt
Now, I = ∫(log⁡x2)/x dx
= ∫(2log⁡x)/x dx
= 2∫(log⁡x)/x dx …….(ii)
Now on putting log⁡x = t and dx/x = dt in equation (ii), we get
I = 2∫tdt
= (2t2)/2 + c
= t+ c
I = (log⁡x)+ c
Question 29. ∫(sin⁡x)/(1 + cos⁡x)2 dx
Solution:
Given that I = ∫(sin⁡x)/((1 + cos⁡x)2) dx …..(i)
Let us considered 1 + cos⁡x = t then,
On differentiating both side we get,
d(1 + cos⁡x) = dt
-sin⁡xdx = dt
sin⁡xdx = -dt
Now on putting 1 + cos⁡x = t and sin⁡dx = -dt in equation (i), we get
I = ∫(-dt)/t2
= -∫t-2dt
= -(-1t-1) + c
= 1/t + c
= 1/(1 + cos⁡x) + c
Hence, I = 1/(1 + cos⁡x) + c
Question 30. ∫cot⁡x log⁡ sin⁡x dx
Solution:
Given that I = ∫cot⁡x log⁡ sin⁡x dx
Let us considered log⁡ sin⁡x = t
1/(sin⁡x).cos⁡xdx = dt
cot⁡x dx = dt
∫cot⁡x log⁡ sin⁡x dx = ∫tdt
= t2/2 + c
= 1/2(log⁡sin⁡x)+ c
Question 31. ∫sec⁡x.log⁡(sec⁡x + tan⁡x)dx
Solution:
Given that I = ∫sec⁡x.log⁡(sec⁡x + tan⁡x)dx  ……..(i)
Let us considered log⁡(sec⁡x + tan⁡x) = t then,
On differentiating both side we get,
d[log⁡(sec⁡x + tan⁡x)] = dt
sec⁡x dx = dt    [Since, d/dx(log⁡(sec⁡x + tan⁡x)) = sec⁡x]
Now on putting log⁡(sec⁡x + tan⁡x) = t and sec⁡x dx = dt in equation (i), we get
I = ∫tdt
= t2/2 + c
= 1/2[log⁡(sec⁡x + tan⁡x)]+ c
Hence, I = 1/2[log⁡(sec⁡x + tan⁡x)]+ c
Question 32. ∫cosec⁡x log⁡(cosec⁡x – cot⁡x)dx
Solution:
Given that I = ∫cosec⁡x log⁡(cosec⁡x – cot⁡x)dx    ……(i)
Let us considered log⁡(cosec⁡x – cot⁡x) = t then,
On differentiating both side we get,
dx[log⁡(cosec⁡x – cot⁡x)] = dt
cosec⁡x dx = dt     [ Since, d/dx(log⁡(cosec⁡x – cot⁡x)) = cosec⁡x]
Now on putting log⁡(cosec⁡x – cot⁡x) = t and cosec⁡xdx = dt in equation (i), we get
I = ∫tdt
= t2/2 + c
Hence, I = 1/2[log⁡(cosec⁡x – cot⁡x)]+ c
Question 33. ∫x3cos⁡x4 dx
Solution:
Given that I = ∫x3cos⁡x4 dx   …….(i)
Let us considered x= t then,
On differentiating both side we get,
dx(x4) = dt
4x3dx = dt
x= dt/4
Now on putting x= t and x3dx = dt/4 in equation (i), we get
I = ∫ cos⁡t dt/4
= 1/4sint + c
Hence, I = 1/4sinx+ c
Question 34. ∫x3 sin⁡x4 dx
Solution:
Given that I = ∫x3sin⁡x4 dx   …..(i)
Let us considered x= t then,
On differentiating both side we get,
d(x4) = dt
4x3dx = dt
x= dt/4
Now on putting x= t and x3dx = dt/4 in equation (i), we get
I = ∫sin⁡t dt/4
= 1/4 ∫sin⁡t dt
= -1/4 cos⁡t + c
Hence, I = -1/4 cos⁡x+ c
Question 35. ∫(xsin-1x2)/√(1 – x4) dx
Solution:
Given that I = ∫(xsin-1⁡x2)/√(1 – x4) dx  …….(i)
Let us considered sin-1x= t then,
On differentiating both side we get,
d(sin-1x2) = dt
2x × 1/√(1 – x4) dx = dt
x/√(1 – x4) dx = dt/2
Now on putting sin-1x= t and x/√(1 – x4) dx = dt/2 in equation (i), we get
I = ∫t dt/2
= 1/2 × t2/2 + c
= 1/4 (sin-1x2)+ c
Hence, I = 1/4 (sin-1x2)+ c
Question 36. ∫x3sin⁡(x+ 1)dx
Solution:
Given that I = ∫x3 sin⁡(x+ 1)dx ……..(i)
Let us considered x+ 1 = t then,
On differentiating both side we get,
d(x+ 1) = dt
x3 dx = dt/4
Now on putting x+ 1 = t and x3dx = dt/4 in equation (i), we get
I = ∫ sin⁡t dt/4
= -1/4 cos⁡t + c
= -1/4 cos⁡(x+ 1) + c
Hence, I = -1/4 cos⁡(x+ 1) + c
Question 37. ∫(x + 1)ex/(cos2⁡(xex) dx
Solution:
Given that I = ∫((x + 1)ex)/(cos2⁡(xex)) dx   ……(i)
Let us considered xe= t then,
On differentiating both side we get,
d(xex) = dt
(e+ xex)dx = dt
(x + 1)exdx = dt
Now on putting xe= t and (x + 1)exdx = dt in equation (i), we get
I = ∫dt/(cos2⁡t)
= ∫ sec2tdt
= tan⁡t + c
= tan⁡(xex) + c
Hence, I = tan⁡(xex) + c
Question 38. Solution:
Given that I = ……..(i)
Let us considered = t then,
On differentiating both side we get,
d( ) = dt
3x2 dx = dt
x2 dx = dt/3
Now on putting = t and x2 dx = dt/3 in equation (i), we get
I = ∫cos⁡t dt/3
= (sin⁡t)/3 + c
Hence, I = sin⁡( )/3 + c
Question 39. ∫2xsec3(x+ 3)tan⁡(x+ 3)dx
Solution:
Given that I = ∫2xsec3(x+ 3)tan⁡(x+ 3)dx    ………(i)
Let us considered sec⁡(x+ 3) = t then,
On differentiating both side we get,
d[sec⁡(x+ 3)] = dt
2xsec⁡(x+ 3)tan⁡(x+ 3)dx = dt
Now on putting sec⁡(x+ 3) = t and 2xsec⁡(x+ 3)tan⁡(x+ 3)dx = dt in equation (i), we get
I = ∫t2 dt
= t3/3 + c
= 1/3 [sec⁡(x+ 3)]+ c
Hence, I = 1/3 [sec⁡(x+ 3)]+ c
Question 40. ∫(1 + 1/x)(x + log⁡x)2 dx
Solution:
Given that I = ((x + 1)(x + log⁡x)2)/x
= ((x + 1)/x)(x + log⁡x)2
= (1 + 1/x)(x + log⁡x)2
Let us considered (x + log⁡x) = t
On differentiating both side we get,
(1 + 1/x)dx = dt
Now,
I = ∫(1 + 1/x)(x + log⁡x)2 dx
= ∫t2 dt
= t3/3 + c
Hence, I = 1/3(x + log⁡x)+ c
Question 41. ∫tan⁡x sec2⁡x√(1 – tan2x) dx
Solution:
Given that I = ∫tan⁡x sec2⁡x√(1 – tan2x) dx      ………(i)
Let us considered 1 – tan2⁡x = t then,
On differentiating both side we get,
d(1 – tan2x) = dt
-2tan⁡x sec2⁡x dx = dt
tan⁡x sec2⁡x dx = (-dt)/2
Now on putting 1 – tan2x = t and tan⁡x sec2⁡x dx = -dt/2 in equation (i), we get
I = ∫√t × (-dt)/2
=-1/2 ∫t1/2 dt
=-1/2×t3/2/(3/2) + c
=-1/3 t3/2 + c
Hence, I = -1/3 [1 – tan2⁡x]3/2 + c
Question 42.∫log⁡x (sin⁡(1 + (log⁡x)2)/x dx
Solution:
Given that I = ∫log⁡x (sin⁡(1 + (log⁡x)2)/x dx  ……..(i)
Now on putting 1 + (log⁡x)= t and (log⁡x)/x dx = dt/2 in equation (i), we get
I = ∫sin⁡t × dt/2
= 1/2 ∫ sin⁡tdt
= -1/2 cos⁡t + c
= -1/2 cos⁡[1 + (log⁡x)2] + c
Hence, I = -1/2 cos⁡[1 + (log⁡x)2] + c
Question 43.∫ 1/x2 × (cos2(1/x))dx
Solution:
Given that I = ∫ 1/x2 × (cos2⁡(1/x))dx ……(i)
Let us considered 1/x = t then,
On differentiating both side we get,
d(1/x) = dt
(-1)/x2dx = dt
1/xdx = -dt
Now on putting 1/x = t and 1/x2dx = -dt in equation (i), we get
I = ∫cos2t(-dt)
= -∫cos2tdt
= -∫(cos2t + 1)/2 dt
= -1/2 ∫cos⁡2t dt – 1/2 ∫dt
= -1/2 × (sin⁡2t)/2 – 1/2 t + c
= -1/4 sin⁡2t – 1/2 t + c
= -1/4 sin⁡2 × 1/x – 1/2 × 1/x + c
Hence, I = -1/4 sin⁡(2/x) – 1/2 (1/x) + c
Question 44. ∫sec4x tan⁡x dx
Solution:
Given that I = ∫sec4x tan⁡x dx  ……(i)
Let us considered tan⁡x = t then,
On differentiating both side we get,
d (tan⁡x) = dt
sec2xdx = dt
dx = dt/sec2⁡x
Now on putting tan⁡x = t and dx = dt/(sec2⁡x) in equation (i), we get
I = ∫sec4x tan⁡x dt/(sec2⁡x)
= ∫ sec2x tdt
= ∫ (1 + tan2⁡x)tdt
= ∫(1 + t2)tdt
= ∫(t + t3)dt
= t2/2 + t4/4 + c
= (tan2⁡x)/2 + (tan4⁡x)/4 + c
Hence, I = 1/2 tan2⁡x + 1/4 tan4⁡x + c
Question 45. ∫(e√x cos⁡(e√x ))/√x dx
Solution:
Given that I = ∫(e√x cos⁡(e√x))/√x dx  …….(i)
Let us considered e√x = t then,
On differentiating both side we get,
d(e√x) = dt
e√x(1/(2√x))dx = dt
e√x/√x dx = 2dt
Now on putting e√x = t and e√x/√x dx = 2dt in equation (i), we get
I = ∫ cos⁡t × 2dt
= 2∫ cos⁡tdt
= 2sin⁡t + c
= 2sin⁡(e√x) + c
I = 2sin⁡(e√x) + c
Question 46. ∫(cos5x)/(sin⁡x) dx
Solution:
Given that I = ∫(cos5⁡x)/(sin⁡x) dx    …..(i)
Let us considered sin⁡x = t then,
On differentiating both side we get,
d(sin⁡x) = dt
cos⁡x dx = dt
dx = dt/(cos⁡x)
Now on putting sin⁡x = t and dx = dt/(cos⁡x) in equation (i), we get
I = ∫(cos5⁡x)/t × dt/(cos⁡x)
= ∫(cos4⁡x)/t dt
= ∫(1 – sin2⁡x)2/t dt
= ∫(1 – t2)2/t dt
= ∫(1 + t– 2t2)/t dt
= ∫1/t dt + ∫t4/t dt – 2∫t2/t dt
= log⁡|t| + t4/4 – (2t2)/2 + c
= log⁡|sin⁡x| + (sin4⁡x)/4 – sin2⁡x + c
Hence, I = 1/4 sin4x – sin2x + log⁡|sin⁡x| + c
Question 47. ∫(sin⁡√x)/√x dx
Solution:
Given that I = ∫(sin⁡√x)/√x dx
Let us considered √x = t then,
On differentiating both side we get,
1/(2√x) dx = dt
1/√x dx = 2dt
Now,
I = ∫(sin⁡√x)/√x dx
= 2 ∫sint dt
= -2 cos⁡t + c
Hence, I = -2cos⁡√x + c
Question 48. ∫((x + 1)ex)/(sin2(xex)) dx
Solution:
Given that I = ∫((x + 1)ex)/(sin2⁡(xex)) dx   …….(i)
Let us considered xe= t then,
d(xex) = dt
(xe+ ex)dx = dt
(x + 1)exdx = dt
Now on putting xe= t and (x + 1)ex dx = dt in equation (i), we get
I = ∫dt/(sin2t)
= ∫cos⁡ec2t dt
= -cot + c
Hence, I = -cot⁡(xex) + c

### Question 49. Solution:

Given that I = ……..(i)

Let us considered x + tan-1⁡x = t then,

On differentiating both side we get,

d(x + tan-1⁡x) = dt

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

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

((x+ 2))/((x+ 1)) dx = dt

Now on putting x + tan-1⁡x = t and ((x+ 2)/(x+ 1))dx = dt in equation (i), we get

I = ∫5tdt

= 5t/(log⁡5) + c /(log⁡5) + c

Hence, I = /(log⁡5) + c

### Question 50. Solution:

Given that I = ……(i)

Let us considered msin-1⁡x = t then,

On differentiating both side we get,

d(msin-1x) = dt

m 1/√(1 – x2) dx = dt

dx/√(1 – x2) = dt/m

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

I = ∫etdt/m

= 1/m et+c Hence, I = ### Question 51. ∫(cos⁡√x)/√x dx

Solution:

Given that I = ∫(cos⁡√x)/√x dx

Let us considered √x = t then,

On differentiating both side we get,

1/(2√x) dx = dt

Now,

= ∫ (cos⁡√x)/√x dx

= 2∫ cos⁡tdt2

= 2sin⁡t + c

Hence, I = 2sin⁡√x + c

### Question 52. ∫sin(tan-1x)/(1 + x2) dx

Solution:

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

Let us considered tan-1 = t then,

On differentiating both side we get,

d(tan-1⁡x) = dt

1/(1 + x2) dx = dt

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

I = ∫ sin⁡tdt

= -cos⁡t + c

= -cos⁡(tan-1x) + c

Hence, I = -cos⁡(tan-1⁡x) + c

### Question 53. ∫(sin⁡(log⁡x))/x dx

Solution:

Given that I = ∫(sin⁡(log⁡x))/x 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 = ∫sin⁡tdt

= -cos⁡t + c

= -cos⁡(log⁡x) + c

Hence, I = -cos⁡(log⁡x) + c

### Question 54. Solution:

Given that I = Let us considered tan-1⁡x = t, then

On differentiating the above function we have,

1/(1 + x2) dx = dt = ∫emt × dt

= = emt/m

On Substituting the value of t, we

I = + c

### Question 55. ∫x/(√(x2 + a2) + √(x2 – a2)) dx

Solution:

Given that I = ∫x/(√(x+ a2) + √(x– a2)) dx

= ∫ x/(√(x2 + a²) + √(x2 – a2)) × (√(x2 + a2) – √(x2 – a2 ))/(√(x+ a2) – √(x2 – a2)) dx

= ∫ x(√(x+ a2) – √(x– a2))/(x+ a– x+ a2) dx

= ∫ x/(2a2) (√(x+ a2) – √(x– a2))dx

I = 1/(2a2) ∫x(√(x+ a2) – √(x– a2))dx ……(i)

Let us considered x= t then,

On differentiating the above function we have,

d(x2) = dt

2xdx = dt

xdx = dt/2

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

I = 1/(2a2) ∫(√(t + a2) – √(t – a2)) dx/2

Hence, I = 1/(4a2) [2/3 (t + a2)3/2 – 2/3 (t – a2)3/2] + c

### Question 56. ∫x(tan-1⁡x2)/(1 + x4) dx

Solution:

Given that I = ∫x(tan-1⁡x2)/(1 + x4) dx  ……..(i)

Let us considered tan-1x= t then,

On differentiating the above function we have,

d(tan-1x2) = dt

(1 × 2x)/(1 + (x2)2) dx = dt

(1 × x)/(1 + x4) dx = dt/2

Now on putting tan-1⁡x= t and x/(1 + x4) dx = dt/2 in equation (i), we get

I = ∫ t dx/2

= 1/2 ∫tdt

= 1/2 × t2/2 + c

= t2/4 + c – 1

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

Hence, I = 1/4 (tan-1⁡x2)+ c

### Question 57. ∫(sin-1x)3/√(1 – x2) dx

Solution:

Given that I = ∫(sin-1x)3/√(1 – x2) dx ……(i)

Let us considered sin-1x = t then,

On differentiating the above function we have,

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 = ∫t3 dt

= t4/4 + c

Hence, I = 1/4 (sin-1x)+ c

### Question 58.∫(sin⁡(2 + 3log⁡x))/x dx

Solution:

Given that I = ∫(sin⁡(2 + 3log⁡x))/x dx  ……..(i)

Let us considered 2 + 3log⁡x = t then,

On differentiating the above function we have,

d(2 + 3log⁡x) = dt

3 1/x dx = dt

dx/x = dt/3

Now on putting 2 + 3log⁡x = t and dx/x = dt/3 in equation (i), we get

I = ∫sin⁡t dt/3

= 1/3(-cos⁡t) + ct

= -1/3 cos⁡(2 + 3log⁡x) + c

Hence, I = -1/3 cos⁡(2 + 3log⁡x) + c

### Question 59. Solution:

Given that I = ……(i)

Let us considered x2 = t then,

On differentiating the above function we have,

d(x2) = dt

2xdx = dt

xdx = dt/2

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

I = ∫etdt/2

= 1/2 et+c

= 1/2 + c

Hence, I = 1/2 + c

### Question 60. ∫e2x/(1 + ex) dx

Solution:

Given that I = ∫e2x/(1 + ex) dx  …….(i)

Let us considered 1 + e= t then,

On differentiating the above function we have,

d(1 + ex) = dt

exdx = dt

dx = dt/ex

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

I = ∫e2x/t × dt/ex

= ∫ex/t dt

= ∫ (t – 1)/t dt

= ∫ (t/t – 1/t)dt

= t – log⁡|t| + c

= (1 + ex) – log⁡|1 + ex| + c

Hence, I = 1 + e– log⁡|1 + ex| + c

### Question 61. ∫(sec2⁡√x)/√x dx

Solution:

Given that I = ∫(sec2√x)/√x dx   ……(i)

Let us considered √x = t then,

On differentiating the above function we have,

d(√x) = dt

1/(2√x) dx = dt

dx = 2√x dt

dx = 2tdt            [√x = t])

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

I = ∫ (sec2t)/t × 2tdt

= 2∫ sec2⁡tdt

= 2tan⁡t + c

= 2tan⁡√x + c

Hence, I = 2tan⁡√x + c

### Question 62. ∫tan32x sec⁡2x dx

Solution:

Given that I = ∫tan32x sec⁡2x dx

= tan22xtan⁡2x sec⁡2x

= (sec22x – 1)tan⁡2x sec⁡2x

= sec22x.tan⁡2xsec⁡2x – tan⁡2xsec⁡2x

= ∫ sec2⁡2xtan⁡2xsec⁡2xdx – ∫tan⁡2xsec⁡2xdx

= ∫ sec22xtan⁡2xsec⁡2xdx – (sec⁡2x)/2 + c

Let us considered sec⁡2x = t

2sec⁡2xtan⁡2xdx = dt

I = 1/2 ∫t2 dt – (sec⁡2x)/2 + c

I = t3/6 – (sec⁡2x)/2 + c

Hence, I = (sec⁡2x)3/6 – (sec⁡2x)/2 + c

### Question 63. ∫(x + √(x + 1))/(x + 2) dx

Solution:

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

Let us considered x + 1 = t2 then,

On differentiating the above function we have,

d(x + 1) = d(t2)

dx = 2tdt

Now on putting x + 1 = t2 and dx = 2tdt in equation (i), we get

I = ∫ (x + √(t2))/(x + 2) 2tdt

= 2∫((t– 1) + t)/((t– 1) + 2) × tdt   [x + 1 = t2]

= 2∫(t+ t – 1)/(t+ 1) tdt

= 2∫ (t+ t– t)/(t2 + 1) dt

= 2[∫ t3/(t+ 1) dt + ∫ t2/(t+ 1) dt – ∫ t/(t+ 1) dt]

I = 2[∫t3/(t+ 1) dt + ∫t2/(t+ 1) dt – ∫t/(t+ 1) dt]    ……(ii)

Let I= ∫t3/(t+ 1) dt

I= ∫t2/(t+ 1) dt

and I= ∫t/(t+ 1) dt

Now, I= ∫t3/(t+ 1) dt

= ∫(t – t/(t+ 1))dt

= t2/2 – 1/2 log⁡(t+ 1)

I= t2/2 – 1/2 log⁡(t+ 1) + c1    ……..(iii)

Since, I= ∫t2/(t+ 1) dt

= ∫ (t+ 1 – 1)/(t+ 1) dt

= ∫(t+ 1)/(t+ 1) dt – ∫1/(t+ 1) dt

= ∫dt – ∫1/(t+ 1) dt

I= t – tan-1⁡(t2) + c2 ………….(iv)

and,

I= ∫t/(t+ 1) dt

= 1/2 log⁡(1 + t2) + c3   ……..(v)

Using equations (ii), (iii), (iv) and (v), we get

I = 2[t2/2 – 1/2 log⁡(t+ 1) + c+ t-tan-1⁡(t2) + c– 1/2 log⁡(1 + t2) + c3]

= 2[t2/2 + t-tan-1⁡(t2) – log⁡(1 + t2) + c+ c+ c3]

= 2[t2/2 + t – tan-1⁡(t2) – log⁡(1 + t2) + c4  [Putting c+ c+ c= c4]

= t+ 2t – 2tan-1(t2) – 2log⁡(1 + t2) + 2c4

= (x + 1) + 2√(x + 1) – 2tan-1(√(x + 1)) – 2log⁡(1 + x + 1) + 2c4

= (x + 1) + 2√(x + 1) – 2tan-1⁡(√(x + 1)) – 2log⁡(x + 2) + c [Putting 2c= c]

Hence, I = (x + 1) + 2√(x + 1) – 2tan-1⁡(√(x + 1)) – 2log⁡(x + 2) + c

### Question 64. Solution:

Given that I = …….(1)

Let us considered = t then

On differentiating the above function we have,

d( ) = dt × × 5x × (log⁡5)3 dx = dt dx = dt/((log⁡5)^3 ))

Now on putting = t and dx = dt/((log⁡5)^3 )) in equation (i), we get

I = ∫dt/((log⁡5)3)

= 1/((log⁡5)3) ∫dt

= t/((log⁡5)^3) + c

Hence, I = /((log⁡5)3) + c)

### Question 65. ∫1/(x√(x4 – 1)) dx

Solution:

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

Let us considered x2 = t then,

On differentiating the above function we have,

d(x2) = dt

2xdx = dt

dx = dt/2x

Now on putting x2 = t and dx = dt/2x in equation (i), we get

I = ∫ 1/(x√(t– 1)) × dt/2x

= 1/2 ∫ 1/(x2 √(t– 1)) dt

= 1/2 ∫ 1/(t√(t– 1)) dt

= 1/2 sec-1t + c

= 1/2 sec-1x+ c

Hence, I = 1/2 sec-1x2 + c

### Question 66. ∫√(ex – 1) dx

Solution:

Given that I = ∫√(e– 1) dx   ……..(i)

Let us considered e– 1 = t2 then,

On differentiating the above function we have,

d(e– 1) = dt(t2)

ex dx = 2tdt

dx = 2t/edt

dx = 2t/(t2 + 1) dt               [e– 1 = t2

Now on putting e– 1 = t² and dx = 2tdt/(t2 + 1) in equation (i), we get

I = ∫√(t2) × 2tdt/(t2 + 1)

= 2∫(t × t)/(t2 + 1) dt

= 2∫t2/(t2 + 1) dt

= 2∫(t2 + 1 – 1)/(t2 + 1) dt

= 2∫[(t2 + 1)/(t+ 1) – 1/(t+ 1)]dt

= 2∫dt – 2∫1/(t2 + 1) dt

= 2t – 2tan-1⁡(t) + c

= 2√(e– 1) – 2tan-1(√(e– 1)) + c

Hence, I = 2√(e– 1) – 2tan-1⁡√(e– 1) + c

### Question 67. ∫ 1/(x + 1)(x2 + 2x + 2) dx

Solution:

Given that I = ∫1/(x + 1)(x+ 2x + 2) dx

= ∫1/(x + 1)((x + 1)+ 1) dx

Let us considered x + 1 = tan u then,                               [tanu = Perpendicular/Base = (x + 1)/1]

On differentiating the above function we have,

dx = sec2u du                                                                   [Hypotenuses = √(x+ 2x + 2)]

I = ∫sec2u/tanu(tan2u + 1) du

= ∫ cosu/sinu du

= log| sinu |+c

= log| sin(x + 1)| + c                                         [As we know, sin(x + 1) = P/H = (x + 1)/√(x+ 2x + 2)]

Hence,  I = log| x + 1/√(x+ 2x + 2)| + c

### Question 68. ∫x5/√(1 + x3) dx

Solution:

Given that I = ∫ x5/(√(1 + x3)) dx   …….(i)

Let us considered 1 + x= t2, then

On differentiating the above function we have,

d(1 + x3) = d(t2)

3x2 dx = 2t * dt

dx = 2t dt/3x2

Now on putting 1 + x= t2 and dx = 2tdt/3x2 in equation (i), we get

I = ∫ x5/√t2 * 2t/3xdt

= ∫x5/t * 2t/3xdt

= 2/3∫x3 dt

= 2/3 ∫( t– 1) dt

= 2/3[t3/3 – 2t/3] + c

Hence, I = 2/9(1 + x3)3/2 – 2 √(1 + x2)/3 + c

### Question 69. ∫4x3 √(5 – x2) dx

Solution:

Given that I = ∫4x3 √(5 – x2) dx  ……(i)

Let us considered 5 – x= tthen,

On differentiating the above function we have,

d(5 – x2) = t2

-2xdx = 2tdt

dx = (-t)/x dt

Now on putting 5 – x= t2 and dx = (-t)/x dt in equation (i), we get

I = ∫4x3 √(t2) × (-t)/x dt

= -4∫ x2 t × tdt

= -4∫(5 – t2) t2 dt         [5 – x= t2]

= -4∫(5t– t5)dt

= -20×t3/3 + 4 t5/5 + c

= (-20)/3 × t+ 4/5 × t+ c

= (-20)/3 × (5 – x2)3/2 + 4/5 × (5 – x2)5/2 + c

I = (-20)/3 × (5 -x2)3/2 + 4/5 × (5 – x2)5/2 + c

### Question 70. ∫1/(√x + x) dx

Solution:

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

Let us considered √x = t then,

On differentiating the above function we have,

d(√x) = dt

1/(2√x) dx = dt

dx = 2√x dt

Now on putting √x = t and 2√x dt = dx in equation (i), we get

I = ∫1/(t + t2) 2t × dt  [Since √x = t and x = t2]

= ∫2t/(t(1 + t)) dt

= 2∫t/(1 + t) dt

= 2log⁡|1 + t| + c

= 2log⁡|1 + √x| + c

Hence, I = 2log⁡|1 + √x|+c

### Question 71. ∫1/(x2 (x4 + 1)3/4) dx

Solution:

Given that I = 1/(x2 (x4+1)3/4)

Multiplying and dividing by x-3, we obtain

(x-3/(x2x-3 (x4+ 1)3/4) = (x-3 (x+ 1)-3/4/(x2x-3))

= (x+ 1)-3/4/(x5(x4)-3/4

= 1/x5 ((x+ 1)/x4)-3/4

= 1/x5 (1 + 1/x4)-3/4

Let us considered 1/x4 = t

-4/xdx = dt

1/xdx = -dt/4

I = ∫ 1/x5(1 + 1/x4)-3/4 dx

= -1/4 ∫(1 + t)-3/4 dt

= -1/4 [(1 + t)1/4)/(1/4)] + c

= -1/4(1 + 1/x4)1/4/(1/4) + c

Hence, I = -(1 + 1/x4)1/4 + c

### Question 72. ∫(sin⁡5x)/(cos4x) dx

Solution:

Given that I = ∫(sin⁡5x)/(cos4⁡x) dx  ……(i)

Let us considered cos⁡x = t then,

On differentiating the above function we have,

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 = ∫(sin5⁡x)/t× -dt/(sin⁡x)

= -∫(sin4⁡x)/t4 dt

= -∫(1 – cos2x)2/tdt

= -∫(1 – t2)2/tdt

= -∫(1 + t– 2t2)/t4dt

= -∫(1/t4 + t4/t4 – (2t2)/t4)dt

= -∫(t-4 + 1 – 2t-2)dt

= -[t-3/(-3) + t – 2 t-1/(-1)] + c

= 1/3 * 1/t3– t – 1/t + c

Hence, I = 1/3 * 1/cos3x – cosx -2/cosx + c

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