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# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.25 | Set 1

• Last Updated : 16 May, 2021

### 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 x4 + 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 = xex – ∫1.ex dx

= xex – ex + c

Hence, I = = xex – ex + 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 = ∫x2 cos⁡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)/xn dx = ∫(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

= tet – ∫et dt

= tet – et + 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 x2 = 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⁡x2 + 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

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