# RD Sharma Class 12 Ex 19.25 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.25 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 19.25 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 19.25 Solutions Chapter 19 Indefinite Integrals

### Question 1. ∫x cos⁡xdx

Solution:

Given that, I = ∫x cos⁡xdx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = x∫cos⁡xdx – ∫(1 × ∫cos⁡xdx)dx + c

= xsin⁡x – ∫sin⁡xdx + c

Hence, I = x sin⁡x + cos⁡x + c

### Question 2. ∫log⁡(x + 1)dx

Solution:

Given that, I = ∫log⁡(x + 1)dx

= ∫1 × log⁡(x + 1)dx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = log⁡(x + 1)∫1dx – ∫(1/(x + 1) × ∫ 1dx)dx + c

= xlog⁡(x + 1) – ∫(x/(x + 1))dx + c

= x log⁡(x + 1) – ∫(1 – 1/(x + 1))dx + c

Hence, I = x log⁡(x + 1) – x + log⁡(x + 1) + c

### Question 3. ∫x3 log⁡xdx

Solution:

Given that, I = ∫ x3 log⁡xdx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = log⁡x ∫x3 dx – ∫(1/x × ∫x3 dx)dx + c

= x4/4 log⁡x – ∫x4/4x dx+c

= x4/4 log⁡x – 1/4∫x3 dx + c

= x4/4 log⁡x – 1/4 ∫x4/4 dx + c

I = x4/4 log⁡x – 1/16 x+ c

### Question 4. ∫xex dx

Solution:

Given that I = ∫xex dx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = xe– ∫1.ex dx

= xe– e+ c

Hence, I = = xe– e+ c

### Question 5. ∫xe2x dx

Solution:

Given that, I = ∫xe2x dx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = x∫e2x dx – ∫(1 × ∫ e2x dx) dx + c

= x∫e2x dx – ∫(1 × ∫e2x dx)dx + c

= (xe2x)/2 – ∫(e2x/2)dx + c

= (xe2x)/2 – e2x/4 + c

Hence, I = (x/2 – 1/4) e2x + c

### Question 6. ∫x2 e-x dx

Solution:

Given that I = ∫x2 e-x dx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = x2 ∫e-x dx – ∫(2x∫e-x dx)dx

= -x2 e-x – ∫(2x)(-e-x)dx

= -x2 e-x + 2∫xe-x dx

= -x2 e-x + 2[x∫e-x dx – ∫(1 × ∫ e-x dx) dx]

= -x2 e-x + 2[x(-e-x) – ∫(-e-x)dx]

= -x2 e-x – 2xe-x + 2∫e-x dx

Hence, I = -x2 e-x – 2xe-x – 2e-x + c

### Question 7.  ∫ x2cos⁡xdx

Solution:

Given that, I = ∫ x2cos⁡xdx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = x2 ∫ cos⁡xdx – ∫(2x)cos⁡xdx)dx

= x2 sin⁡x – 2∫(x)(sin⁡x)dx

= x2 sin⁡x – 2[x∫sin⁡xdx – ∫(1 × ∫sin⁡xdx)dx]

= x2 sin⁡x – 2[x(-cos⁡x) – ∫(-cos⁡x)dx]

= x2 sin⁡x + 2xcos⁡x – 2∫(cos⁡x)dx

Hence, I = x2sin⁡x + 2xcos⁡x – 2sin⁡x + c

### Question 8. ∫x2cos⁡2xdx

Solution:

Given that, I = ∫x2cos⁡2xdx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = x2 ∫cos⁡2xdx – ∫(2x∫ cos⁡2xdx)dx

= x2 (sin⁡2x)/2 – 2∫x((sin⁡2x)/2)dx

= 1/2 x2 sin⁡2x – ∫xsin⁡2xdx

= 1/2 x2 sin⁡2x – [x∫sin⁡2xdx – ∫ (1∫ sin⁡2xdx)dx]

= 1/2 x2 sin⁡2x – [x((-cos⁡2x)/2) – ∫(-(cos⁡2x)/2)dx]

= 1/2 x2sin⁡2x + x/2 cos⁡2x – 1/2 ∫(cos⁡2x)dx

Hence, I = 1/2 x2 sin⁡2x + x/2 cos⁡2x – 1/4 sin⁡2x + c

### Question 9. ∫xsin⁡2xdx

Solution:

Given that, I =∫xsin⁡2xdx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = x∫sin⁡2xdx – ∫(1)sin⁡2xdx)dx

= x(-(cos⁡2x)/2) – ∫(-(cos⁡2x)/2)dx

= -x/2 cos⁡2x + 1/2 ∫cos⁡2xdx

= -x/2 cos⁡2x + 1/2(sin⁡2x)/2 + c

Hence, I = -x/2 cos⁡2x + 1/4 sin⁡2x + c

### Question 10. ∫(log⁡(log⁡x))/x dx

Solution:

Given that, I = ∫(log⁡(log⁡x))/x dx

= ∫(1/x)(log⁡(log⁡x))dx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = log⁡log⁡x]1/x dx – ∫(1/(xlog⁡x)∫1/x dx)dx

= log⁡x × log⁡(log⁡x) – ∫(1/(xlog⁡x) log⁡x)dx

= log⁡x × log⁡(log⁡x) – ∫1/x dx

= log⁡x × log⁡(log⁡x) – log⁡x + c

Hence, I = log⁡x(log⁡log⁡x – 1) + c

### Question 11. ∫x2 cos⁡xdx

Solution:

Given that I = ∫xcos⁡xdx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = x2∫ cos⁡xdx – ∫(2x]cos⁡xdx)dx

= x2sin⁡x – 2∫xsin⁡xdx

= x2 sin⁡x – 2[x∫sin⁡xdx – ∫(1]sin⁡xdx)dx]

= x2 sin⁡x – 2[x(-cos⁡x) – ∫(-cos⁡x)dx]

= x2 sin⁡x + 2xcos⁡x – 2∫(cos⁡x)dx

Hence, I = x2 sin⁡x + 2xcos⁡x – 2sin⁡x + c

### Question 12. ∫xcosec2⁡xdx

Solution :

Given that, I = ∫xcosec2⁡xdx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = x∫cosec2xdx – ∫(∫ cosec2xdx)dx

= -xcot⁡x + ∫cot⁡xdx

= -x cot⁡x + log ⁡|sin⁡x| + c

Hence, I = -x cot⁡x + log ⁡|sin⁡x| + c

### Question 13. ∫xcos2xdx

Solution:

Given that, I = ∫xcos2xdx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = x∫cos2⁡xdx – ∫(1∫ cos2xdx)dx

= x∫((cos⁡2x + 1)/2)dx – ∫(∫((1 + cos⁡2x)/2)dx)dx

= x/2 [(sin⁡2x)/2 + x] – 1/2∫(x + (sin⁡2x)/2)dx

= x/4 sin⁡2x + x2/2 – 1/2 × x2/2 – 1/4 (-(cos⁡2x)/2) + c

Hence, I = x/4 sin⁡2x + x2/4 + 1/8 cos⁡2x + c

### Question 14. ∫xn log⁡x dx

Solution:

Given that, I = ∫xn log⁡xdx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = log⁡x∫xn dx – ∫(1/x ∫xndx)dx

= xn+1/(n + 1) log⁡x – ∫(1/x × xn+1/(n + 1))dx

= xn+1/(n + 1) log⁡x – ∫(xn/(n + 1))dx

Hence, I = xn+1/(n + 1) log⁡x – 1/(n + 1)2 × (xn+1) + c

### Question 15. ∫(log⁡x)/xn dx

Solution:

Given that, I = ∫(log⁡x)/xdx = ∫(log⁡x)(1/xn)dx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = log⁡x∫(1/xn)dx – ∫((d(log⁡x))/dx)(∫(1/xn)dx)dx

= log⁡x(x1-n/(1 – n)) – ∫1/x (x1-n/(1 – n))dx

= log⁡x(x1-n/(1 – n)) – ∫(xn/(1 – n))dx

= log⁡x(x1-n/(1 – n)) – (1/(1 – n))(x1-n/(1 – n))

Hence, I = log⁡x(x1-n/(1 – n)) – (x1-n/([1 – n]2)) + c

### Question 16. ∫x2 sin2⁡xdx

Solution:

Given that, I = ∫x2 sin2⁡xdx

= ∫x2 ((1 – cos⁡2x)/2)dx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= ∫x2/2 dx – ∫((x2 cos⁡2x)/2)dx

= x3/6 – 1/2 [∫x2 cos⁡2xdx]

= x3/6 – 1/2 [x2 ∫cos⁡2xdx – ∫ (2x∫cos⁡2xdx)dx]

= x3/6 – 1/2 (x2(sin⁡2x)/2) + 1/2 × 2∫(x (sin⁡2x)/2)dx

= x3/6 – 1/4 x2sin⁡2x + 1/2 [x ∫sin⁡2xdx – ∫(1∫sin⁡2xdx)dx]

= x3/6 – 1/4 x2 sin⁡2x + 1/2 [x(-(cos⁡2x)/2) – ∫(-(cos⁡2x)/2)dx]

= x3/6 – 1/4 x2 sin⁡2x + 1/2 x(-(cos⁡2x)/2) + 1/4 × (sin2x/2) + c

= x3/6 – 1/4 x2 sin⁡2x – 1/4 x(cos⁡2x) + 1/8 × (sin2x) + c

Hence, I = x3/6 – 1/4 x2 sin⁡2x – 1/4 x(cos⁡2x) + 1/8 × (sin2x) + c

### Question 17.

Solution:

Given that, l =

Let us assume, x2 = t

2xdx = dt

I = ∫t × et dt

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= t∫et dt – ∫(1 × ∫etdt)dt

= te– ∫et dt

= te– e+ c

= et-1 + c

Hence, I =  (x2 – 1) + c

### Question 18. ∫x3 cos⁡x2 dx

Solution:

Given that, I = ∫x3 cos⁡x2 dx

Let us assume x= t

2xdx = dt

I = 1/2 ∫tcos⁡tdt

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= 1/2[t∫cos⁡tdt – ∫(1 × ∫cos⁡tdt)dt]

= 1/2 [t × sin⁡t – ∫sin⁡tdt]

= 1/2[tsin⁡t + cos⁡t] + c

Hence, I = 1/2 [x² sin⁡x+ cos⁡x2] + c

### Question 19. ∫xsin⁡xcos⁡xdx

Solution:

Given that, I = ∫xsin⁡xcos⁡xdx

= ∫x/2(2sin⁡xcos⁡x)dx

= 1/2 ∫xsin⁡2xdx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= 1/2 [x∫sin⁡2xdx – ∫(1 × ∫sin⁡2xdx)dx]

= 1/2 [x((-cos⁡2x)/2) – ∫((-cos⁡2x)/2)dx]

= -1/4 xcos⁡2x + 1/4 ∫cos⁡2xdx

Hence, I = -1/4 xcos⁡2x + 1/8 sin⁡2x + c

### Question 20. ∫sin⁡x(log⁡cos⁡x)dx

Solution:

Given that, I = ∫sin⁡x(log⁡cos⁡x)dx

Let us considered, cos⁡x = t

-sin⁡xdx = dt

I = -∫ log⁡tdt

= -∫1 × log⁡tdt

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= -[log⁡t∫dt – ∫(1/t × ∫dt)dt]

= -[tlog⁡t – ∫1/t × tdt]

= -[tlog⁡t-∫  dt]

= -[tlog⁡t – t + c1 ]

= t(1 – logt) + c

Hence, I = cosx(1 – logcosx) + c

### Question 21. ∫(log⁡x)2 x dx

Solution:

Given that, I = ∫(log⁡x)2 x dx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = (log⁡x)2∫xdx – ∫(2(log⁡x)(1/x) ∫xdx)dx

= x2/2(log⁡x)– 2∫(log⁡x)(1/x)(x2/2)dx

= x2/2(log⁡x)– ∫x(log⁡x)dx

= x2/2(log⁡x)– [log⁡x∫xdx – ∫ (1/x ∫xdx)dx]

= x2/2(log⁡x)– [x22/2 log⁡x – ∫(1/x × x2/2)dx]

= x2/2(log⁡x)– x2/2 log⁡x + 1/2 ∫xdx

= x2/2(log⁡x)– x2/2 log⁡x + 1/4 x+ c

Hence, I = x2/2 [(log⁡x)– log⁡x + 1/2] + c

### Question 22. ∫e√x dx

Solution:

Given that, I = ∫ e√x dx

Let us assume, √x = t

x = t2

dx = 2tdt

I = 2∫ et tdt

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = 2[t∫etdt – ∫(1∫etdt)dt]

= 2[te– ∫et dt]

= 2[te– et] + c

= 2et (t – 1) + c

Hence, I = 2e√x(√x – 1) + c

### Question 23. ∫(log⁡(x + 2))/((x + 2)2) dx

Solution:

Given that, I = ∫(log⁡(x + 2))/((x + 2)2) dx

Let us assume (1/(x + 2) = t

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

I = -∫log⁡(1/t)dt

= -∫log⁡t-1 dt

= -∫1 × log⁡tdt

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = log⁡t∫dt – ∫(1/t ∫dt)dt

= tlog⁡t – ∫(1/t × t)dt

= tlog⁡t – ∫dt

= tlog⁡t – t + c

= 1/(x + 2) (log⁡(x + 2)-1 – 1) + c

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

### Question 24. ∫(x + sin⁡x)/(1 + cos⁡x) dx

Solution:

Given that, I = ∫(x + sin⁡x)/(1 + cos⁡x) dx

= ∫x/(2cos2x/2) dx + ∫(2sin⁡x/2 cos⁡x/2)/(2cos2⁡x/2) dx

= 1/2 ∫xsec2⁡x/2 dx + ∫tan⁡x/2 dx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= 1/2 [x∫sec2x/2 dx – ∫(1∫ sec²x/2 dx)dx] + ∫tan⁡x/2 dx

= 1/2 [2xtan⁡x/2 – 2∫tan⁡x/2 dx] + ∫tan⁡x/2 dx + c

= xtan⁡x/2 – ∫tan⁡x/2 dx + ∫tan⁡x/2 dx+c

Hence, I = xtan⁡x/2 + c

### Question 25. ∫log10xdx

Solution:

Given that, I = ∫log10⁡xdx

= ∫(log⁡x)/(log⁡10) dx

= 1/(log⁡10) ∫1 × log⁡xdx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= 1/(log⁡10) [log⁡x∫dx – ∫(1/x ∫dx)dx]

= 1/(log⁡10) [xlog⁡x – ∫(x/x)dx]

= 1/(log⁡10)[xlog⁡x – x]

Hence, I = (x/(log⁡10)) × (log⁡x – 1) + c

### Question 26. ∫cos⁡√x dx

Solution:

Given that, I = ∫cos⁡√x dx

Let us assume, √x = t

x = t2

dx = 2tdt

= ∫2tcos⁡tdt

I = 2∫tcos⁡tdt

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = 2[t]cos⁡tdt – ∫(1 ∫cos⁡tdt)dt]

= 2[tsin⁡t – ∫sin⁡tdt]

= 2[tsin⁡t + cos⁡t] + c

Hence, I = 2[√x sin⁡√x + cos√x] + c

### Question 27. ∫(xcos-1x)/√(1 – x2) dx

Solution:

Given that, I = ∫(xcos-1x)/√(1 – x2) dx

Let us assume, t = cos-1⁡x

dt = (-1)/√(1 – x2) dx

Also, cost = x

I = -∫tcos⁡tdt

Now, using integration by parts,

So, let

u = t;

du = dt

∫cos⁡tdt = ∫dv

sin⁡t = v

Therefore,

I = -[tsint – ∫sin⁡tdt]

= -[tsint + cos⁡t] + c

On substituting the value t = cos-1x we get,

I = -[cos-1⁡xsin⁡t + x] + c

Hence, I = -[cos-1⁡x√(1 – x²) + x] + c

### Question 28. ∫cosec3xdx

Solution:

Given that, I =∫cosec3xdx

=∫cosec⁡x × cosec2xdx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= cosec⁡x × ∫cosec2xdx + ∫(cosec⁡xcot⁡x]cosec2xdx)dx

= cosecx × (-cot⁡x) + ∫cosec⁡xcot⁡x(-cot⁡x)dx

= -cosec⁡xcot⁡x – ∫cosecx⁡cot2⁡xdx

= -cosec⁡xcot⁡x – ∫cosec⁡x(cosec2x – 1)dx

= -cosec⁡xcot⁡x – ∫cosec3xdx + ∫cosecxd⁡x

I = -cosec⁡xcot⁡x – I + log⁡|tan⁡x/2| + c1

2l = -cosec⁡xcot⁡x + log⁡|tan⁡x/2| + c1

Hence, I = -1/2cosecx⁡cot⁡x + 1/2 log⁡|tan⁡x/2| + c

### Question 29. ∫sec-1√x dx

Solution:

Given that, I = ∫sec-1⁡√x dx

Let us assume, √x = t

x = t2

dx = 2tdt

I = ∫2tsec-1⁡tdt

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= 2[sec-1⁡t∫tdt – ∫(1/(t√(t2-1))∫tdt)dt]

= 2[t2/2 sec-1 – ∫(t/(2t√(t– 1)))dt]

= t2 sec-1⁡t – ∫t/√(t– 1) dt

= t2 sec-1⁡t – 1/2∫2t/√(t– 1) dt

= t2 sec-1⁡t – 1/2 × 2√(t– 1) + c

Hence, I = xsec-1⁡√x – √(x – 1) + c

### Question 30. ∫sin-1√x dx

Solution:

Given that, I = ∫sin-1⁡√x dx

Let us assume, x = t

dx = 2tdt

∫sin-1√x dx = ∫sin-1⁡√(t2) 2tdt

= ∫sin-1⁡t2tdt

= sin⁡-1t∫2tdt – (∫(dsin-1⁡t)/dt (∫2tdt)dt

= sin-1⁡t(t2) – ∫1/√(1 – t2) (t2)dt

Now, lets solve ∫1/√(1 – t2) (t2)dt

∫1/√(1 – t2) (t2)dt = ∫(t– 1 + 1)/√(1 – t2) dt

= ∫(t– 1)/√(1 – t2) dt + ∫1/√(1 – t2) dt

As we know that, value of ∫1/√(1 – t2) dt = sin-1⁡t

So, the remaining integral to evaluate is

∫(t– 1)/√(1 – t2) dt= ∫-√(1 – t2) dt

Now, substitute, t = sin⁡u, dt = cos⁡udu, we gte

∫-√(1 – t2) dt = ∫-cos2udu = -∫[(1 + cos⁡2u)/2]du

= -u/2-(sin⁡2u)/4

Now substitute back u = sin-1x and t = √x, we get

= -(sin-1√x)/2 – (sin⁡(2sin-1⁡√x))/4

∫sin-1√x dx = xsin-1⁡√x-(sin-1⁡√x)/2 – (sin⁡(2sin-1√x))/4

sin⁡(2sin-1⁡√x) = 2√x √(1 – x)

Hence, I = xsin-1⁡√x – (sin-1⁡√x)/2 – √(x(1 – x))/2 + c

### Question 31. ∫xtan2⁡xdx

Solution:

Given that, I =∫xtan2⁡xdx

= ∫x(sec2x – 1)dx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= ∫xsec8xdx – ∫xdx

= [x∫sec2xdx – ∫(1∫sec2xdx)dx] – x2/2

= xtan⁡x – ∫tan⁡xdx – x2/2

Hence, I = xtan⁡x – log⁡|sec⁡x| – x2/2 + c

### Question 32. ∫ x((sec⁡2x – 1)/(sec⁡2x + 1))dx

Solution:

Given that, I = ∫ x((sec⁡2x – 1)/(sec⁡2x + 1))dx

= ∫x((1 – cos⁡2x)/(1 + cos⁡2x))dx

= ∫x((sec2⁡x)/(cos2⁡x))dx

= ∫xtan2xdx

= ∫x(sec2⁡x – 1)dx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= ∫xsec2xdx – ∫dx

= [x∫sec2⁡xdx – ∫(1∫  sec2xdx)dx] – x2/2

= xtan⁡x – ∫tan⁡xdx – x2/2

= xtan⁡x – log|secx| – x2/2 + c

Hence, I = xtan⁡x – log|secx| – x2/2 + c

### Question 33. ∫(x + 1)exlog⁡(xex)dx

Solution:

Given that, I = ∫(x + 1)exlog⁡(xex)dx

Let us assume, xe= t

(1 × e+ xex)dx = dt

(x + 1)exdx = dt

I = ∫log⁡tdt

= ∫1 × log⁡tdt

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= log⁡t∫dt – ∫(1/t∫dt)dt

= tlog⁡t – ∫(1/t × t)dt

= tlog⁡t – ∫dt

= tlog⁡t – t + c

= t(log⁡t – 1) + c

Hence, I = xex (log⁡xe– 1) + c

### Question 34. ∫sin-1(3x – 4x3)dx

Solution:

Given that, I = ∫sin-1(3x – 4x3)dx

Let us assume, x = sin⁡θ

dx = cos⁡θdθ

= ∫sin-1⁡(3sin⁡θ – 4sin3⁡θ)cos⁡θdθ

= ∫sin-1(sin⁡3θ)cos⁡θdθ

= ∫3θcos⁡θdθ

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= 3[θ]cos⁡θdθ – ∫(1∫cos⁡θdθ)dθ]

= 3[θsin⁡θ – ∫sin⁡θdθ]

= 3[θsin⁡θ + cos⁡θ] + c

Hence, I = 3[xsin-1⁡x + √(1 – x2)] + c)

### Question 35. ∫sin-1(2x/(1 + x2))dx.

Solution:

Given that, I = ∫sin-1(2x/(1 + x2))dx

Let us assume, x = tan⁡θ

dx = sec2⁡θdθ

sin-1⁡(2x/(1 + x2)) = sin-1⁡((2tan⁡θ)/(1 + tan²⁡θ))

= sin-1⁡(sin⁡2θ) = 2θ

∫sin-1⁡(2x/(1 + x2))dx = ∫2θsec2θdθ = 2∫θsec2⁡θdθ

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

2[θ∫sec2⁡θdθ – ∫{(d/dθ θ) ∫sec2θdθ}dθ

= 2[θtan⁡θ – ∫tan⁡θdθ]

= 2[θtan⁡θ + log⁡|cos⁡θ|] + c

= 2[xtan-1⁡x + log⁡|1/√(1 + x2)|] + c

= 2xtan-1⁡x + 2log⁡(1 + x2)1/2 + c

= 2xtan-1⁡x + 2[-1/2 log⁡(1 + x2)] + c

= 2xtan-1⁡x – log⁡(1 + x2) + c

Hence, I = 2xtan-1⁡x – log⁡(1 + x2) + c

### Question 36. ∫ tan-1((3x – x3)/(1 – 3x2))dx

Solution:

Given that, I = ∫tan-1⁡((3x – x3)/(1 – 3x2))dx

Let us assume, x = tan⁡θ

dx = sec2⁡θdθ

I = ∫tan-1⁡((3tan⁡θ – tan3θ)/(1 – 3tan2⁡x)) sec2⁡θdθ

=∫tan-1⁡(tan⁡3θ)sec2θdθ

= ∫3θsec2θdθ

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= 3[θ∫ sec2⁡θdθ – ∫(1∫sec2θdθ)dθ]

= 3[θtan⁡θ – ∫tan⁡θdθ]

= 3[θtan⁡θ + log⁡sec⁡θ] + c

= 3[xtan-1x – log⁡√(1 + x2)] + c

Hence, I = 3[xtan-1x – log⁡√(1 + x2)] + c

### Question 37. ∫x2sin-1xdx

Solution:

Given that, I = ∫x2sin-1⁡xdx

I = sin-1x∫x2 dx – ∫(1/√(1 – x2) ∫x2 dx)dx

= x3/3 sin-1⁡x – ∫x3/(3√(1 – x2)) dx

I = x3/3 sin-1⁡x – 1/3 I+ c…..(1)

Let I= ∫x3/√(1 – x2) dx

Let 1 – x= t2

-2xdx = 2tdt

-xdx = tdt

I= -∫(1 – t2)tdt/t

= ∫(t– 1)dt

= t3/3 – t + c2

= (1 – x2)3/2/3 – (1 – x2)1/2 + c2

Now, put the value of I1 in eq(1), we get

Hence, I = x3/3 sin-1⁡x – 1/9 (1 – x2)3/2 + 1/3 (1 – x2)1/2 + c

### Question 38. ∫(sin-1x)/x2dx

Solution:

Given that, I =∫(sin-1⁡x)/x2dx

= ∫(1/x2)(sin-1⁡x)dx

I = [sin-1⁡x∫1/x2dx – ∫(1/√(1 – x2) ∫1/x2dx)dx]

= sin-1×(-1/x) – ∫1/√(1 – x2) (-1/x)dx

I = -1/x sin-1x + ∫1/(x√(1 – x2)) dx

I = -1/x sin-1⁡x + I1  …….(1)

Where,

I= ∫1/(x√(1 – x2)) dx

Let 1 – x= t2

-2xdx = 2tdt

I= ∫x/(x2√(1 – x2)) dx

= -∫tdt/((1 – t2) √t)

= -∫dt/((1 – t2))

= ∫1/(t– 1) dt

= 1/2 log⁡|(t – 1)/(t + 1)|

= 1/2 log⁡|(t – 1)/(t + 1)|

= 1/2 log⁡|(√(1 – x2) – 1)/(√(1 – x2) + 1)| + c1

Now, put the value of I1 in eq(1), we get

I = -(sin-1x)/x + 1/2 log⁡|((√(1 – x2) – 1)/(√(1 – x2) + 1))((√(1 – x2) – 1)/(√(1 – x2) – 1))| + c

= -(sin-1⁡x)/x + 1/2 log⁡|(√(1 – x2) – 1)2/(1 – x– 1)| + c

= -(sin-1⁡x)/x + 1/2 log⁡|(√(1 – x2) – 1)2/(-x2)| + c

= -(sin-1⁡x)/x + log⁡|(√(1 – x2) – 1)/(-x)| + c

Hence, I = -(sin-1x)/x + log⁡|(1 – √(1 – x2))/x| + c

### Question 39. ∫(x2 tan-1x)/(1 + x2) dx

Solution:

Given that, I = ∫(x2 tan-1⁡x)/(1 + x2) dx

Let us assume, tan-1⁡x = t     [x = tan⁡t]

1/(1 + x2) dx = dt

I = ∫t × tan2tdt

= ∫t(sec2⁡t – 1)dt

= ∫(tsec2t – t)dt

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= ∫tsec2tdt – ∫tdt

= [t∫sec2tdt – ∫(1)sec2⁡tdt)dt] – t2/2

= [t × tan⁡t – ∫tan⁡tdt] – t2/2

= t tan⁡t – log⁡sec⁡t – t2/2 + c

= xtan-1⁡x – log⁡√(1 + x2) – (tan2x)/2 + c

Hence, I = xtan-1⁡x – 1/2 log⁡|1 + x2| – (tan2⁡x)/2 + c

### Question 40. ∫cos-1(4x3 – 3x)dx

Solution:

Given that, I = ∫cos-1⁡(4x– 3x)dx

Let us assume, x = cos⁡θ

dx = -sin⁡θdθ

I = -∫cos-1(4cos⁡3θ – 3cos⁡θ)sin⁡θdθ

= – ∫cos-1(cos⁡3θ)sin⁡θdθ

= -∫3θsin⁡θdθ

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= -3[θ]sin⁡θdθ – ∫(1∫sin⁡θdθ)dθ]

= -3[-θcos⁡θ + ∫cos⁡θdθ]

= 3θcos⁡θ – 3sin⁡θ + c

Hence, I = 3xcos-1x – 3√(1 – x2) + c

### Question 41. ∫cos-1⁡((1 – x2)/(1 + x2))dx

Solution:

Given that, I = ∫cos-1⁡((1 – x2)/(1 + x2))dx)

Let us considered x = tan⁡t

dx = sec²tdt

I = ∫cos-1⁡((1 – tan2t)/(1 + tan2⁡t)) sec2tdt

= ∫cos-1(cos⁡2t)sec2tdt

= ∫2tsec2⁡tdt

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = 2[t∫sce2tdt – ∫(1∫sec2⁡tdt)dt]

= 2[t × tan2t – ∫tan⁡tdt]

= 2[t × tan2t – log⁡sec⁡t] + c

= 2[xtan-1x – log⁡√(1 + x2)] + c

Hence, I = 2xtan-1x – log⁡|1 + x2| + c

### Question 42. ∫tan-1⁡(2x/(1 – x2))dx

Solution:

Given that, I = ∫tan-1⁡(2x/(1 – x2))dx

Let us considered x = tan⁡θ

dx = sec2θdθ

I = ∫tan-1⁡((2tan⁡θ)/(1 – tan2θ)) sec2θdθ

= ∫tan-1⁡(tan⁡2θ)sec2θdθ

= ∫2θsec2θdθ

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = 2[θ∫sec2θdθ – ∫(1∫ sec2⁡θdθ)dθ]

= 2[θtan⁡θ – ∫tan⁡θdθ]

= 2[θtan⁡θ – log⁡sec⁡θ] + c

= 2[xtan-1⁡x – log⁡√(1 + x2)] + c

Hence, I = 2xtan-1⁡x – log⁡|1 + x2| + c

### Question 43. ∫(x + 1)log⁡xdx

Solution:

Given that, I = ∫(x + 1)log⁡xdx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = log⁡x∫ (x + 1)dx – ∫(1/x ∫(x + 1)dx)dx

= (x2/2 + x)log⁡x – ∫1/x (x2/2 + x)dx

= (x2/2 + x)log⁡x – 1/2 ∫xdx – ∫dx

= (x + x2/2)log⁡x – 1/2 × x2/2 – x + c

Hence, I = (x + x2/2)log⁡x – 1/2 × x2/2 – x + c

### Question 44. ∫ x2 tan-1xdx

Solution:

Given that, I = ∫ x2 tan-1⁡xdx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = tan-1⁡x∫x2 dx – ∫(1/(1 + x2) ∫x2 dx) dx

= tan-1⁡x(x3/3) – 1/3∫x3/(1 + x2) dx

= 1/3 x3 tan-1⁡x – 1/3 ∫(x – x/(1 + x2))dx

= 1/3 x3 tan-1⁡x – 1/3 × x2/2 + 1/3 ∫x/(1 + x2) dx

Hence, I = 1/3 x3tan-1x – 1/6 x+ 1/6 log⁡|1 + x2| + c

### Question 45. ∫(elogx + sin⁡x) cos⁡xdx

Solution:

Given that, I = ∫(elogx + sin⁡x)cos⁡xdx

= ∫(x + sin⁡x)cos⁡xdx

= ∫xcos⁡xdx + ∫sin⁡xcos⁡xdx

= [x∫cos⁡xdx – ∫(1]cos⁡xdx)dx] + 1/2 ∫sin⁡2xdx

= [xsin⁡x – ∫ sin⁡xdx] + 1/2 (-(cos⁡2x)/2) + c

I = xsin⁡x+cos⁡x – 1/4 cos⁡2x + c

= xsin⁡x + cos⁡x – 1/4 [1 – 2sin2⁡x] + c

= xsin⁡x + cos⁡x – 1/4 + 1/2 sin2x + c

= xsin⁡x + cos⁡x – 1/4 + 1/2 sin2x + c

Hence, I = xsin⁡x + cos⁡x + 1/2 sin2⁡x + d   [d = c-/4]

### Question 46. ∫((xtan-1⁡x))/(1 + x2)3/2 dx

Solution:

Given that, I = ∫((xtan-1⁡x))/(1 + x2)3/2dx

Let us considered tan-1⁡x = t

1/(1 + x2) dx = dt

I = ∫(t tan⁡t)/√(1 + tan2⁡t) dt

= ∫(t × tan⁡t)/(sec⁡t) dt

= ∫t (sin⁡t)/(cos⁡t) cos⁡tdt

= ∫tsin⁡tdt

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = [t]sin⁡tdt – ∫(1)sin⁡tdt)dt]

= [-tcos⁡t + ∫cos⁡tdt]

= [-tcos⁡t + sin⁡t] + c

= -(tan-1⁡x)/√(1 + x2) + x/√(1 + x2) + c

Hence, I = -(tan-1⁡x)/√(1 + x2) + x/√(1 + x2) + c

### Question 47. ∫ tan-1(√x)dx

Solution:

Given that, I = ∫ tan-1(√x)dx

Let us considered x = t2

dx = 2tdt

I = ∫2ttan-1tdt

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= 2[tan-1)⁡t∫tdt – ∫(1/(1 + t2) ∫tdt)dt]

= 2[t2/2 tan-1⁡t – ∫t2/2(1 + t2)dt]

= t2 tan-1⁡t – ∫(t2 + 1 – 1)/(1 + t2)dt

= t2 tan-1⁡t – ∫(1 – 1/(1 + t2))dt

= t2 tan-1t – t + tan-1⁡t + c

= (t2 + 1) tan-1⁡t – t + c

Hence, I = (x + 1)tan-1⁡√x – √x + c

### Question 48. ∫x3 tan-1xdx

Solution:

Given that, I = ∫x3 tan-1xdx

= tan-1⁡x∫x3dx – (∫(dtan-1⁡x)/dx (∫x3 dx)dx)

= tan-1⁡x x4/4 – (∫1/(1 + x2) (x4/4)dx)

= tan-1⁡x x4/4 – (∫1/(1 + x2) (x4/4)dx)

= tan-1⁡x x4/4 – (∫1/(1 + x2) (x4/4)dx)

∫ 1/(1 + x2) (x4/4)dx = 1/4 [∫1/(1 + x2) dx + (x2 – 1)dx]

∫ 1/(1 + x2) (x4/4)dx = 1/4 [tan-1⁡x + x3/3 – x]

Hence, I = x4/4 tan-1⁡x – 1/4 [tan-1⁡x + x3/3 – x] + c

### Question 49. ∫xsin⁡xcos⁡2xdx

Solution:

Given that, I = ∫xsin⁡xcos⁡2xdx

= 1/2 ∫x(2sin⁡xcos⁡2x)dx

= 1/2 ∫x(sin⁡(x + 2x) – sin⁡(2x – x))dx

= 1/2 ∫x(sin⁡3x – sin⁡x)dx

= 1/2[x](sin⁡3x – sin⁡x)dx – ∫ (1)(sin⁡3x – sin⁡x)dx)dx]

= 1/2 [x((-cos⁡3x)/3 + cos⁡x) – ∫(-(cos⁡3x)/3 + cos⁡x)dx]

Hence, I = 1/2 [-x (cos⁡3x)/3 + xcos⁡x + 1/9 sin⁡3x – sin⁡x] + c

### Question 50. ∫(tan-1x2)xdx

Solution:

Given that, I = ∫(tan-1⁡x2)xdx

Let us considered x2 = t

2xdx = dt

I = 1/2∫tan-1tdt

= 1/2∫1tan-1tdt

= 1/2 [tan-1⁡t∫dt – (∫1/(1 + t2)∫dt)dt]

= 1/2 [t × tan-1⁡t – ∫t/(1 + t2) dt]

= 1/2 t × tan-1⁡t – 1/4∫2t/(1 + t2) dt

= 1/2 t × tan-1⁡t – 1/4 log⁡|1 + t2| + c

Hence, I = 1/2 x2 tan-1⁡x2 – 1/4 log⁡|1 + x4| + c

### Question 51. ∫xdx/√(1 – x2)

Solution:

Given that, I = ∫xdx/√(1 – x2)

Let first function be sin-1⁡x and second function be x/√(1 – x2).

Now, first we find the integral of the second function,

∫xdx/√(1 – x2)

Now, put t = 1 – x2

Then dt = -2xdx

Therefore,

∫ xdx/√(1 – x2) = -1/2 ∫dt/√t = -√t = -√(1 – x2)

Hence,

∫(xsin-1x)/√(1 – x2) dx

= (sin-1⁡x)(-√(1 – x2) – ∫1/√(1 – x2) * (-√(1 – x2))dx

= -√(1 – x2) sin-1⁡x + x + c

= x – √(1 – x2) sin-1⁡x + c

### Question 52. ∫sin3√x dx

Solution:

Given that, I = ∫sin3√x dx

Let us considered √x = t

x = t2

dx = 2tdt

I = 2∫ tsin3⁡tdt

= 2∫t((3sin⁡t – sin⁡3t)/4)dt

= 1/2 ∫t(3sin⁡t – sin⁡3t)dt

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = 1/2 [t(-3cos⁡t + 1/3 cos⁡3t) – ∫(-3cos⁡t + (cos⁡3t)/3)dt]

= 1/2 [(-9tcos⁡t + tcos⁡3t)/3 – {-3sin⁡t + (sin⁡3t)/9}] + c

= 1/2 [(-9tcos⁡t + tcos⁡3t)/3 + (27sin⁡t – 3sin⁡3t)/9] + c

= 1/18[-27tcos⁡t + 3tcos⁡3t + 27sin⁡t – 3sin⁡3t] + c

Hence, I = 1/18[3√x cos⁡3√x + 27sin⁡√x – 27√x cos⁡√x – 3sin⁡3√x] + c

### Question 53. ∫ xsin3xdx

Solution:

Given that, I = ∫ xsin3⁡xdx

= ∫x((3sin⁡x – sin⁡3x)/4)dx

= 1/4 ∫x(3sin⁡x – sin⁡3x)dx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= 1/4 [x∫ (3sin⁡x – sin⁡3x)dx – ∫(1)(3sin⁡x – sin⁡3x)dx)dx]

= 1/4 [x(-3cos⁡x + (cos⁡3x)/3) – ∫(-3cos⁡x + (cos⁡3x)/3)dx]

= 1/4 [-3xcos⁡x + (xcos⁡3x)/3 + 3sin⁡x – (sin⁡3x)/9] + c

Hence, I = 1/36[3xcos⁡3x – 27xcos⁡x + 27sin⁡x – sin⁡3x] + c

### Question 54. ∫cos3√x dx

Solution:

Given that, I = ∫cos3√x dx

Let us considered x = t²

dx = 2tdt

= 2∫tcos3⁡tdt

= 2∫t((3cos⁡t + cos⁡3t)/4)dt

= 1/2 ∫t(3cos⁡t + cos⁡3t)dt

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = 1/2 [t(3sin⁡t + 1/3 sin⁡3t) + ∫(1 × 3sin⁡t + (sin⁡3t)/3)dt]

= 1/2 [t((9sin⁡t + sin⁡3t)/3) + 3cos⁡t(cos⁡3t)/9] + c

= 1/18[27tsin⁡t + 3tsin⁡3t + 9cos⁡t + cos⁡3t] + c

Hence, I = 1/18[27√x sin⁡√x + 3√x sin⁡3√x + 9cos⁡√x + cos⁡3√x] + c

### Question 55. ∫xcos3xdx

Solution:

Given that, I = ∫xcos3⁡xdx

= ∫x((3cos⁡x + cos⁡3x)/4)dx

= 1/4 ∫x(3cos⁡x + cos⁡3x)dx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = 1/4 [x∫(3cos⁡x + cos⁡3x)dx – ∫(1)(3cos⁡x + cos⁡3x)dx)dx]

= 1/4 [x(3sin⁡x + (sin⁡3x)/3) – ∫ (3sin⁡x + (sin⁡3x)/3)dx]

= 1/4 [3xsin⁡x + (xsin⁡3x)/3 + 3cos⁡x + (cos⁡3x)/9] + c

Hence, I = (3xsin⁡x)/4 + (xsin⁡3x)/12 + (3cos⁡x)/4 + (cos⁡3x)/36 + c

### Question 56. ∫tan-1√((1 – x)/(1 + x))

Solution:

Given that, I = ∫tan-1√((1 – x)/(1 + x))

Let us considered x = cos⁡θ

dx = -sin⁡θdθ

I = ∫ tan-1⁡(tan⁡θ/2)(-sin⁡θ)dθ

=-1/2 ∫θsin⁡θdθ

Let θ = u and sin⁡θdθ = v

So that sin⁡θ = ∫vdθ

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = -1/2 (-θcos⁡θ – ∫-cos⁡θdθ)

= -1/2(-θcos⁡θ + sin⁡θ)+c

= -1/2 (-θcos⁡θ + √(1 – cos2⁡θ)) + c

= -1/2 (-xcos-1⁡x + √(1 – x2)) + c

### Question 57. ∫sin-1√(x/(a + x)) dx

Solution:

Given that, I = ∫sin-1⁡√(x/(a + x)) dx

Let us considered x = atan2θ

dx = 2atan⁡θsec2⁡θdθ

I = ∫(sin-1⁡√((atan2⁡θ)/(a + atan2⁡θ))(2atan⁡θsec2θ)dθ

= ∫ (sin-1√((tan2θ)/(sec2θ)))(2atan⁡θsec2θ)dθ

= ∫ sin-1(sin⁡θ)(2atan⁡θsec2θ)dθ

= ∫ 2θatan⁡θsec2θdθ

= 2a∣θ(tan⁡θsec2⁡θ)dθ)

= ∫2θatan⁡θsec2θdθ

= 2a∫θ(tan⁡θsec2⁡θ)dθ

= 2a[θ]tan⁡θsec2θdθ – ∫(∫tan⁡θsec2⁡θdθ)dθ]

= 2a[θ (tan2⁡θ)/2 – ∫(tan2θ)/2 dθ]

= aθtan2θ – 2a/2∫(sec2θ – 1)dθ

= aθtan2θ – atan⁡θ + aθ + c

= a(tan-1⁡√(x/a)) x/a – a√(x/a) + atan-1⁡√(x/a) + c

Hence, I = xtan-1⁡√(x/a) – √ax + atan-1⁡√(x/a) + c

### Question 58. ∫(x3 sin-1⁡x²)/√(1 – x4) dx

Solution:

Given that, I = ∫(x3 sin-1x²)/√(1 – x4) dx

Let us considered sin-1⁡x² = t

(1/√(1 – x4)(2x)dx = dt

I = ∫(x² sin-1⁡x²)/√(1 – x4) xdx

= ∫(sin⁡t)t dt/2

= 1/2∫tsin⁡tdt

= 1/2 [t∫sin⁡tdt – ∫(1∫sin⁡tdt)dt]

= 1/2 [t(-cost)dt – ∫(1∫(-cost))dt]

= 1/2[-tcost + sint] + c

Hence, I = 1/2 [x2 – √(1 – x4) sin(-1)⁡x2] + c

### Question 59. ∫(x2 sin-1⁡x)/(1 – x2)3/2 dx

Solution:

Given that, I = ∫(x2 sin-1x)/(1 – x2)3/2dx

Let us considered sin-1⁡x = t

(1/√(1 – x2) dx = dt

I = ∫(sin2t × t)/((1 – sin2t)) dt

= ∫(tsin2t)/(cos2t) dt

= ∫t × tan2tdt

= ∫t(sec2⁡t – 1)dt

= ∫tsec2⁡tdt – t2/2 + c

= t∫sec2tdt – ∫(1∫sec2tdt)dt – t2/2 + c

= t × tan⁡t – ∫tan⁡tdt – t2/2 + c

= t × tan⁡t – log⁡sec⁡t – t2/2 + c

Hence, I = x/√(1 – x2) sin-1x + log⁡|1 – x2| – 1/2 (sin-1x)2 + c

### Question 60. ∫cos-1(1 – x2/ 1 + x2) dx

Solution:

Given that, I = ∫cos-1(1 – x2/ 1 + x2) dx

Let us considered, x = tant

dx = sec2tdt

I = ∫cos-1(1 – tan2t/ 1 + tan2t) sec2tdt

= ∫ 2t sec2tdt

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = 2[t∫sec2tdt – ∫(1 ∫sec2tdt)dt]

= 2[t tan2t – ∫tant dt]

= 2[t tan2t – log sect] + c

= 2[x tan2x – log √1 + x2] + c

Hence, I = 2[xtan2x – log √1 + x2] + c

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