# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.25 | Set 2

• Last Updated : 16 Dec, 2021

### Question 21. ā«(logā”x)2 x dx

Solution:

Given that, I = ā«(logā”x)2 x dx

Using integration by parts,

ā«u v dx = vā« u dx – ā«{d/dx(v) Ć ā«u dx}dx + c

We get

I = (logā”x)2ā«xdx – ā«(2(logā”x)(1/x) ā«xdx)dx

= x2/2(logā”x)2 – 2ā«(logā”x)(1/x)(x2/2)dx

= x2/2(logā”x)2 – ā«x(logā”x)dx

= x2/2(logā”x)2 – [logā”xā«xdx – ā« (1/x ā«xdx)dx]

= x2/2(logā”x)2 – [x22/2 logā”x – ā«(1/x Ć x2/2)dx]

= x2/2(logā”x)2 – x2/2 logā”x + 1/2 ā«xdx

= x2/2(logā”x)2 – x2/2 logā”x + 1/4 x2 + c

Hence, I = x2/2 [(logā”x)2 – logā”x + 1/2] + c

### Question 22. ā«eāx dx

Solution:

Given that, I = ā« eāx dx

Let us assume, āx = t

x = t2

dx = 2tdt

I = 2ā« et tdt

Using integration by parts,

ā«u v dx = vā« u dx – ā«{d/dx(v) Ć ā«u dx}dx + c

We get

I = 2[tā«etdt – ā«(1ā«etdt)dt]

= 2[tet – ā«et dt]

= 2[tet – et] + c

= 2et (t – 1) + c

Hence, I = 2eāx(āx – 1) + c

### Question 23. ā«(logā”(x + 2))/((x + 2)2) dx

Solution:

Given that, I = ā«(logā”(x + 2))/((x + 2)2) dx

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

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

I = -ā«logā”(1/t)dt

= -ā«logā”t-1 dt

= -ā«1 Ć logā”tdt

Using integration by parts,

ā«u v dx = vā« u dx – ā«{d/dx(v) Ć ā«u dx}dx + c

We get

I = logā”tā«dt – ā«(1/t ā«dt)dt

= tlogā”t – ā«(1/t Ć t)dt

= tlogā”t – ā«dt

= tlogā”t – t + c

= 1/(x + 2) (logā”(x + 2)-1 – 1) + c

Hence, I = (-1)/(x + 2) – (logā”(x + 2))/(x + 2) + c

### Question 24. ā«(x + sinā”x)/(1 + cosā”x) dx

Solution:

Given that, I = ā«(x + sinā”x)/(1 + cosā”x) dx

= ā«x/(2cos2x/2) dx + ā«(2sinā”x/2 cosā”x/2)/(2cos2ā”x/2) dx

= 1/2 ā«xsec2ā”x/2 dx + ā«tanā”x/2 dx

Using integration by parts,

ā«u v dx = vā« u dx – ā«{d/dx(v) Ć ā«u dx}dx + c

We get

= 1/2 [xā«sec2x/2 dx – ā«(1ā« secĀ²x/2 dx)dx] + ā«tanā”x/2 dx

= 1/2 [2xtanā”x/2 – 2ā«tanā”x/2 dx] + ā«tanā”x/2 dx + c

= xtanā”x/2 – ā«tanā”x/2 dx + ā«tanā”x/2 dx+c

Hence, I = xtanā”x/2 + c

### Question 25. ā«log10xdx

Solution:

Given that, I = ā«log10ā”xdx

= ā«(logā”x)/(logā”10) dx

= 1/(logā”10) ā«1 Ć logā”xdx

Using integration by parts,

ā«u v dx = vā« u dx – ā«{d/dx(v) Ć ā«u dx}dx + c

We get

= 1/(logā”10) [logā”xā«dx – ā«(1/x ā«dx)dx]

= 1/(logā”10) [xlogā”x – ā«(x/x)dx]

= 1/(logā”10)[xlogā”x – x]

Hence, I = (x/(logā”10)) Ć (logā”x – 1) + c

### Question 26. ā«cosā”āx dx

Solution:

Given that, I = ā«cosā”āx dx

Let us assume, āx = t

x = t2

dx = 2tdt

= ā«2tcosā”tdt

I = 2ā«tcosā”tdt

Using integration by parts,

ā«u v dx = vā« u dx – ā«{d/dx(v) Ć ā«u dx}dx + c

We get

I = 2[t]cosā”tdt – ā«(1 ā«cosā”tdt)dt]

= 2[tsinā”t – ā«sinā”tdt]

= 2[tsinā”t + cosā”t] + c

Hence, I = 2[āx sinā”āx + cosāx] + c

### Question 27. ā«(xcos-1x)/ā(1 – x2) dx

Solution:

Given that, I = ā«(xcos-1x)/ā(1 – x2) dx

Let us assume, t = cos-1ā”x

dt = (-1)/ā(1 – x2) dx

Also, cost = x

I = -ā«tcosā”tdt

Now, using integration by parts,

So, let

u = t;

du = dt

ā«cosā”tdt = ā«dv

sinā”t = v

Therefore,

I = -[tsint – ā«sinā”tdt]

= -[tsint + cosā”t] + c

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

I = -[cos-1ā”xsinā”t + x] + c

Hence, I = -[cos-1ā”xā(1 – xĀ²) + x] + c

### Question 28. ā«cosec3xdx

Solution:

Given that, I =ā«cosec3xdx

=ā«cosecā”x Ć cosec2xdx

Using integration by parts,

ā«u v dx = vā« u dx – ā«{d/dx(v) Ć ā«u dx}dx + c

We get

= cosecā”x Ć ā«cosec2xdx + ā«(cosecā”xcotā”x]cosec2xdx)dx

= cosecx Ć (-cotā”x) + ā«cosecā”xcotā”x(-cotā”x)dx

= -cosecā”xcotā”x – ā«cosecxā”cot2ā”xdx

= -cosecā”xcotā”x – ā«cosecā”x(cosec2x – 1)dx

= -cosecā”xcotā”x – ā«cosec3xdx + ā«cosecxdā”x

I = -cosecā”xcotā”x – I + logā”|tanā”x/2| + c1

2l = -cosecā”xcotā”x + logā”|tanā”x/2| + c1

Hence, I = -1/2cosecxā”cotā”x + 1/2 logā”|tanā”x/2| + c

### Question 29. ā«sec-1āx dx

Solution:

Given that, I = ā«sec-1ā”āx dx

Let us assume, āx = t

x = t2

dx = 2tdt

I = ā«2tsec-1ā”tdt

Using integration by parts,

ā«u v dx = vā« u dx – ā«{d/dx(v) Ć ā«u dx}dx + c

We get

= 2[sec-1ā”tā«tdt – ā«(1/(tā(t2-1))ā«tdt)dt]

= 2[t2/2 sec-1 – ā«(t/(2tā(t2 – 1)))dt]

= t2 sec-1ā”t – ā«t/ā(t2 – 1) dt

= t2 sec-1ā”t – 1/2ā«2t/ā(t2 – 1) dt

= t2 sec-1ā”t – 1/2 Ć 2ā(t2 – 1) + c

Hence, I = xsec-1ā”āx – ā(x – 1) + c

### Question 30. ā«sin-1āx dx

Solution:

Given that, I = ā«sin-1ā”āx dx

Let us assume, x = t

dx = 2tdt

ā«sin-1āx dx = ā«sin-1ā”ā(t2) 2tdt

= ā«sin-1ā”t2tdt

= sinā”-1tā«2tdt – (ā«(dsin-1ā”t)/dt (ā«2tdt)dt

= sin-1ā”t(t2) – ā«1/ā(1 – t2) (t2)dt

Now, lets solve ā«1/ā(1 – t2) (t2)dt

ā«1/ā(1 – t2) (t2)dt = ā«(t2 – 1 + 1)/ā(1 – t2) dt

= ā«(t2 – 1)/ā(1 – t2) dt + ā«1/ā(1 – t2) dt

As we know that, value of ā«1/ā(1 – t2) dt = sin-1ā”t

So, the remaining integral to evaluate is

ā«(t2 – 1)/ā(1 – t2) dt= ā«-ā(1 – t2) dt

Now, substitute, t = sinā”u, dt = cosā”udu, we gte

ā«-ā(1 – t2) dt = ā«-cos2udu = -ā«[(1 + cosā”2u)/2]du

= -u/2-(sinā”2u)/4

Now substitute back u = sin-1x and t = āx, we get

= -(sin-1āx)/2 – (sinā”(2sin-1ā”āx))/4

ā«sin-1āx dx = xsin-1ā”āx-(sin-1ā”āx)/2 – (sinā”(2sin-1āx))/4

sinā”(2sin-1ā”āx) = 2āx ā(1 – x)

Hence, I = xsin-1ā”āx – (sin-1ā”āx)/2 – ā(x(1 – x))/2 + c

### Question 31. ā«xtan2ā”xdx

Solution:

Given that, I =ā«xtan2ā”xdx

= ā«x(sec2x – 1)dx

Using integration by parts,

ā«u v dx = vā« u dx – ā«{d/dx(v) Ć ā«u dx}dx + c

We get

= ā«xsec8xdx – ā«xdx

= [xā«sec2xdx – ā«(1ā«sec2xdx)dx] – x2/2

= xtanā”x – ā«tanā”xdx – x2/2

Hence, I = xtanā”x – logā”|secā”x| – x2/2 + c

### Question 32. ā« x((secā”2x – 1)/(secā”2x + 1))dx

Solution:

Given that, I = ā« x((secā”2x – 1)/(secā”2x + 1))dx

= ā«x((1 – cosā”2x)/(1 + cosā”2x))dx

= ā«x((sec2ā”x)/(cos2ā”x))dx

= ā«xtan2xdx

= ā«x(sec2ā”x – 1)dx

Using integration by parts,

ā«u v dx = vā« u dx – ā«{d/dx(v) Ć ā«u dx}dx + c

We get

= ā«xsec2xdx – ā«dx

= [xā«sec2ā”xdx – ā«(1ā«  sec2xdx)dx] – x2/2

= xtanā”x – ā«tanā”xdx – x2/2

= xtanā”x – log|secx| – x2/2 + c

Hence, I = xtanā”x – log|secx| – x2/2 + c

### Question 33. ā«(x + 1)exlogā”(xex)dx

Solution:

Given that, I = ā«(x + 1)exlogā”(xex)dx

Let us assume, xex = t

(1 Ć ex + xex)dx = dt

(x + 1)exdx = dt

I = ā«logā”tdt

= ā«1 Ć logā”tdt

Using integration by parts,

ā«u v dx = vā« u dx – ā«{d/dx(v) Ć ā«u dx}dx + c

We get

= logā”tā«dt – ā«(1/tā«dt)dt

= tlogā”t – ā«(1/t Ć t)dt

= tlogā”t – ā«dt

= tlogā”t – t + c

= t(logā”t – 1) + c

Hence, I = xex (logā”xex – 1) + c

### Question 34. ā«sin-1(3x – 4x3)dx

Solution:

Given that, I = ā«sin-1(3x – 4x3)dx

Let us assume, x = sinā”Īø

dx = cosā”ĪødĪø

= ā«sin-1ā”(3sinā”Īø – 4sin3ā”Īø)cosā”ĪødĪø

= ā«sin-1(sinā”3Īø)cosā”ĪødĪø

= ā«3Īøcosā”ĪødĪø

Using integration by parts,

ā«u v dx = vā« u dx – ā«{d/dx(v) Ć ā«u dx}dx + c

We get

= 3[Īø]cosā”ĪødĪø – ā«(1ā«cosā”ĪødĪø)dĪø]

= 3[Īøsinā”Īø – ā«sinā”ĪødĪø]

= 3[Īøsinā”Īø + cosā”Īø] + c

Hence, I = 3[xsin-1ā”x + ā(1 – x2)] + c)

### Question 35. ā«sin-1(2x/(1 + x2))dx.

Solution:

Given that, I = ā«sin-1(2x/(1 + x2))dx

Let us assume, x = tanā”Īø

dx = sec2ā”ĪødĪø

sin-1ā”(2x/(1 + x2)) = sin-1ā”((2tanā”Īø)/(1 + tanĀ²ā”Īø))

= sin-1ā”(sinā”2Īø) = 2Īø

ā«sin-1ā”(2x/(1 + x2))dx = ā«2Īøsec2ĪødĪø = 2ā«Īøsec2ā”ĪødĪø

Using integration by parts,

ā«u v dx = vā« u dx – ā«{d/dx(v) Ć ā«u dx}dx + c

We get

2[Īøā«sec2ā”ĪødĪø – ā«{(d/dĪø Īø) ā«sec2ĪødĪø}dĪø

= 2[Īøtanā”Īø – ā«tanā”ĪødĪø]

= 2[Īøtanā”Īø + logā”|cosā”Īø|] + c

= 2[xtan-1ā”x + logā”|1/ā(1 + x2)|] + c

= 2xtan-1ā”x + 2logā”(1 + x2)1/2 + c

= 2xtan-1ā”x + 2[-1/2 logā”(1 + x2)] + c

= 2xtan-1ā”x – logā”(1 + x2) + c

Hence, I = 2xtan-1ā”x – logā”(1 + x2) + c

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

Solution:

Given that, I = ā«tan-1ā”((3x – x3)/(1 – 3x2))dx

Let us assume, x = tanā”Īø

dx = sec2ā”ĪødĪø

I = ā«tan-1ā”((3tanā”Īø – tan3Īø)/(1 – 3tan2ā”x)) sec2ā”ĪødĪø

=ā«tan-1ā”(tanā”3Īø)sec2ĪødĪø

= ā«3Īøsec2ĪødĪø

Using integration by parts,

ā«u v dx = vā« u dx – ā«{d/dx(v) Ć ā«u dx}dx + c

We get

= 3[Īøā« sec2ā”ĪødĪø – ā«(1ā«sec2ĪødĪø)dĪø]

= 3[Īøtanā”Īø – ā«tanā”ĪødĪø]

= 3[Īøtanā”Īø + logā”secā”Īø] + c

= 3[xtan-1x – logā”ā(1 + x2)] + c

Hence, I = 3[xtan-1x – logā”ā(1 + x2)] + c

### Question 37. ā«x2sin-1xdx

Solution:

Given that, I = ā«x2sin-1ā”xdx

I = sin-1xā«x2 dx – ā«(1/ā(1 – x2) ā«x2 dx)dx

= x3/3 sin-1ā”x – ā«x3/(3ā(1 – x2)) dx

I = x3/3 sin-1ā”x – 1/3 I1 + c1 …..(1)

Let I1 = ā«x3/ā(1 – x2) dx

Let 1 – x2 = t2

-2xdx = 2tdt

-xdx = tdt

I1 = -ā«(1 – t2)tdt/t

= ā«(t2 – 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-1ā”x – 1/9 (1 – x2)3/2 + 1/3 (1 – x2)1/2 + c

### Question 38. ā«(sin-1x)/x2dx

Solution:

Given that, I =ā«(sin-1ā”x)/x2dx

= ā«(1/x2)(sin-1ā”x)dx

I = [sin-1ā”xā«1/x2dx – ā«(1/ā(1 – x2) ā«1/x2dx)dx]

= sin-1Ć(-1/x) – ā«1/ā(1 – x2) (-1/x)dx

I = -1/x sin-1x + ā«1/(xā(1 – x2)) dx

I = -1/x sin-1ā”x + I1  …….(1)

Where,

I1 = ā«1/(xā(1 – x2)) dx

Let 1 – x2 = t2

-2xdx = 2tdt

I1 = ā«x/(x2ā(1 – x2)) dx

= -ā«tdt/((1 – t2) āt)

= -ā«dt/((1 – t2))

= ā«1/(t2 – 1) dt

= 1/2 logā”|(t – 1)/(t + 1)|

= 1/2 logā”|(t – 1)/(t + 1)|

= 1/2 logā”|(ā(1 – x2) – 1)/(ā(1 – x2) + 1)| + c1

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

I = -(sin-1x)/x + 1/2 logā”|((ā(1 – x2) – 1)/(ā(1 – x2) + 1))((ā(1 – x2) – 1)/(ā(1 – x2) – 1))| + c

= -(sin-1ā”x)/x + 1/2 logā”|(ā(1 – x2) – 1)2/(1 – x2 – 1)| + c

= -(sin-1ā”x)/x + 1/2 logā”|(ā(1 – x2) – 1)2/(-x2)| + c

= -(sin-1ā”x)/x + logā”|(ā(1 – x2) – 1)/(-x)| + c

Hence, I = -(sin-1x)/x + logā”|(1 – ā(1 – x2))/x| + c

### Question 39. ā«(x2 tan-1x)/(1 + x2) dx

Solution:

Given that, I = ā«(x2 tan-1ā”x)/(1 + x2) dx

Let us assume, tan-1ā”x = t     [x = tanā”t]

1/(1 + x2) dx = dt

I = ā«t Ć tan2tdt

= ā«t(sec2ā”t – 1)dt

= ā«(tsec2t – t)dt

Using integration by parts,

ā«u v dx = vā« u dx – ā«{d/dx(v) Ć ā«u dx}dx + c

We get

= ā«tsec2tdt – ā«tdt

= [tā«sec2tdt – ā«(1)sec2ā”tdt)dt] – t2/2

= [t Ć tanā”t – ā«tanā”tdt] – t2/2

= t tanā”t – logā”secā”t – t2/2 + c

= xtan-1ā”x – logā”ā(1 + x2) – (tan2x)/2 + c

Hence, I = xtan-1ā”x – 1/2 logā”|1 + x2| – (tan2ā”x)/2 + c

### Question 40. ā«cos-1(4x3 – 3x)dx

Solution:

Given that, I = ā«cos-1ā”(4x3 – 3x)dx

Let us assume, x = cosā”Īø

dx = -sinā”ĪødĪø

I = -ā«cos-1(4cosā”3Īø – 3cosā”Īø)sinā”ĪødĪø

= – ā«cos-1(cosā”3Īø)sinā”ĪødĪø

= -ā«3Īøsinā”ĪødĪø

Using integration by parts,

ā«u v dx = vā« u dx – ā«{d/dx(v) Ć ā«u dx}dx + c

We get

= -3[Īø]sinā”ĪødĪø – ā«(1ā«sinā”ĪødĪø)dĪø]

= -3[-Īøcosā”Īø + ā«cosā”ĪødĪø]

= 3Īøcosā”Īø – 3sinā”Īø + c

Hence, I = 3xcos-1x – 3ā(1 – x2) + c

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