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

Evaluate the following 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 x⁴ + c

Question 4. ∫xe^x dx

Solution:

Given that I = ∫xe^x dx

Using integration by parts,       

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

We get

I = xe^x - ∫1.e^x dx

= xe^x - e^x + c

Hence, I = = xe^x - e^x + c

Question 5. ∫xe^2x dx

Solution:

Given that, I = ∫xe^2x dx

Using integration by parts,       

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

We get

I = x∫e^2x dx - ∫(1 × ∫ e^2x dx) dx + c

= x∫e^2x dx - ∫(1 × ∫e^2x dx)dx + c

= (xe^2x)/2 - ∫(e^2x/2)dx + c

= (xe^2x)/2 - e^2x/4 + c

Hence, I = (x/2 - 1/4) e^2x + 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(x^1-n/(1 - n)) - ∫1/x (x^1-n/(1 - n))dx

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

= log⁡x(x^1-n/(1 - n)) - (1/(1 - n))(x^1-n/(1 - n))

Hence, I = log⁡x(x^1-n/(1 - n)) - (x^1-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. ∫2x^3 e^{x^2} xdx

Solution:

Given that, l = ∫2x^3 e^{x^2} xdx

 Let us assume, x² = t

2xdx = dt

I = ∫t × e^t dt

Using integration by parts,       

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

We get

= t∫e^t dt - ∫(1 × ∫e^tdt)dt

= te^t - ∫e^t dt

= te^t - e^t + c

= e^t-1 + c

Hence, I = e^{x^2} (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 

Summary

Exercise 19.25 | Set 1 typically deals with integrating rational functions where the denominator is of the form 1 ± x⁴. Key points to remember:

These integrals often require partial fraction decomposition.

• For 1 - x⁴, it can be factored as (1 - x²)(1 + x²) or (1 - x)(1 + x)(1 + x²).
• For 1 + x⁴, it can be factored as (1 + x²)² - (√2x)² = (1 + √2x + x²)(1 - √2x + x²).
• After partial fraction decomposition, you'll usually end up with simpler rational terms.
• These terms can be integrated using standard integration techniques or by recognizing standard forms.