Class 12 RD Sharma Solutions Chapter 19 Indefinite Integrals Exercise 19.11 (original) (raw)

Last Updated : 3 Sep, 2024

Question 1. Integrate ∫ tan3 x sec2 x dx

**Solution:

Let, I = ∫ tan3 x sec2 x dx .......... (i)

Put,

tan x = t .............. (ii)

d(tanx) = dt

sec2 x dx = dt .............. (iiI)

Put equ (ii) and (iii) in equ (i), we get

= ∫ t3 dt

Integrate the above equation, we get

= t3+1 /3+1 + c

= t4 / 4 + c

Put the value of t from equ (ii), we get

= (tan x)4 / 4 + c

Hence, I = tan4 x / 4 + c

Question 2. Integrate ∫ tanx sec4 x dx

**Solution:

Let, I = ∫ tan x sec4 x dx

We can write the above equation as below,

= ∫ tanx sec2 x sec2 x dx

= ∫ tanx (1+tan2 x) sec2 x dx

= ∫ (tanx + tan3x) sec2x dx .......... (i)

Substituting,

tanx = t ........... (ii)

d(tanx) = dt

sec2 dx = dt ........... (iii)

put equ (ii) and (iii) in equ (i), we get

= ∫ (t + t3) dt

Integrate the above equation, we get

= t2/2 + t3+1/3+1 + c

= t2/2 + t4/4 + c

put the value of t in the above equ then, we get

= (tanx)2/2 + (tanx)4 / 4 + c

Hence, I = tan2 x/2 + tan4 x / 4 + c

Question 3. Integrate ∫ tan5 x sec4 x dx

**Solution:

Let, I = ∫ tan5 x sec4 x dx

On solving the above equation,

= ∫ tan4 x sec2 x sec2 xdx

= ∫ tan4 x (1+ tan2 x) sec2 xdx

= ∫ (tan5 x + tan7 x) sec2 xdx ............ (i)

Put, tan x = t ............ (ii)

sec2 x dx = dt ........... (iii)

Put equ (ii) and (iii) in equ (i), we get

= ∫ t5 + t7 dt

Integrate the above equation, we get

= t6 /6 + t8 /8 + c

= (tan x)6/6 + (tan x)8/8 + c

Hence, I = tan6 x/6 + tan8 x/8 + c

Question 4. Integrate ∫ sec6 x tanx dx

**Solution:

Let, I = ∫ sec6 x tanx dx

On solving the above equation,

= ∫ sec5 x (sec x tanx) dx

Put, sec x = t

sec x tan x dx = dt

= ∫ t5 dt

Integrate the above equation, we get

= t6 /6 + c

= (sec x)6 + c

Hence, I = sec6 x/6 + c

Question 5. Integrate ∫ tan5 x dx

**Solution:

Let, I = ∫ tan5 x dx

We can modify the above equation as below,

= ∫ tan2 x tan3 x dx

= ∫ (sec2 x - 1) tan3 x dx

= ∫ sec2 x tan3 x - tan3 x dx

= ∫ sec2 x tan3 x dx - ∫tan3 x dx

= ∫ sec2 x tan3 x dx - (∫(sec2 x - 1) tan x dx)

= ∫ sec2 x tan3 x dx - ∫ sec2 x tan x dx + ∫ tan x dx

Put, tan x = t

sec2 x dx = dt

= ∫ t3 dt - ∫ t dt + ∫ tan x dx

Now, Integrate the above equation

= t4 /4 - t2 /2 + log |sec x| + c

= (tanx)4 /4 - (tanx)2 /2 + log |sec x| + c

Hence, I = tan4 x/4 - tan2 x /2 + log |sec x| + c

Question 6. Integrate ∫ √tanx.sec4 x dx

**Solution:

Let, I = ∫ √tanx.sec4 x dx

= ∫ √tanx.sec2 x sec2 x dx

= ∫ √tanx.(1 + tan2 x) sec2 x dx

= ∫ tan1/2 x.(1 + tan2 x) sec2 x dx

= ∫ (tan1/2 x + tan5/2 x) sec2 x dx

Put, tan x = t

sec2 x dx = dt

= ∫ (t1/2 + t5/2) dt

Now, Integrate the above equation,

= t3/2/(3/2) + t7/2/(7/2) + c

= (2/3) t3/2 + (2/7) t7/2 + c

= (2/3) (tanx)3/2 + (2/7) (tanx)7/2 + c

Hence, I = (2/3) tan3/2 x + (2/7) tan7/2 x + c

Question 7. Integrate ∫ sec4 2x dx

**Solution:

Let, I = ∫ sec4 2x dx

= ∫ sec2 2x . sec2 2x dx

= ∫(1 + tan2 2x) sec2 2x dx

= ∫ sec2 2x + sec2 2x. tan2 2x dx

= ∫ sec2 2x. tan2 2x dx + ∫ sec2 2x dx

Put, tan 2x = t

sec2 2x dx = dt/2

= ∫ t2 dt/2 + ∫ sec2 2x dx

Integrate the above equation then, we get

= 1/2 × t3 /3 + 1/2 tan 2x + c

= 1/6 (tan2x)3 + 1/2 tan 2x + c

Hence, I = = 1/2 tan2x + 1/6 tan3 2x + c

Question 8. Integrate ∫ cosec4 3x dx

**Solution:

Let, I = ∫ cosec4 3x dx

= ∫ cosec2 3x cosec2 3x dx

= ∫ (1+ cot2 3x) cosec2 3x dx

= ∫ (cosec2 3x + cosec2 3x cot2 3x) dx

= ∫ cosec2 3x dx +∫ cosec2 3x cot2 3x dx

Put, cot 3x = t

cosec2 3x dx = - dt/3

= ∫ cosec2 3x dx - ∫ t2 dt/3

Integrate the above equation then, we get

= -cot 3x/3 - t3/9 + c

= (-1/3) cot 3x - (cot 3x)3/9 + c

Hence, I = (-1/3) cot 3x - (1/9) cot3 3x + c

Question 9. Integrate ∫ cotn x cosec2 x dx (n ≠ -1)

**Solution:

Let, I = ∫ cotn x cosec2 x dx, (n≠ -1)

Put, cot x = t

- cosec2 x dx = dt

= - ∫ tn dt

Integrate the above equation, we get

= - tn+1 /(n+1) + c

= - (cot x)n+1 / (n+1) + c

Hence, I = - cotn+1 x/ (n+1) + c

Question 10. Integrate ∫ cot5 x cosec4 x dx

**Solution:

Let, I = ∫ cot5 x cosec4 x dx

= ∫ cot5 x cosec2 x cosec2 x dx

= ∫ cot5 x (1+ cot2 x) cosec2 x dx

= ∫ (cot5 x + cot7 x) cosec2 x dx

Put, cot x = t

- cosec2 x dx = dt

= - ∫ (t5 + t7) dt

Integrate the above equation, we get

= - t6 /6 - t8/8 + c

= -(cotx)6/6 - (cotx)8/8 + c

Hence, I = -cot6 x/6 - cot8 x/8 + c

Question 11. Integrate ∫ cot5 x dx

**Solution:

Let, I = ∫ cot5 x dx

We can modify the above equation as below,

= ∫ cot3 x cot2 x dx

= ∫ cot3 x (cosec2 x - 1) dx

= ∫ cot3 x cosec2 x - cot3 x dx

= ∫ cot3 x cosec2 x dx - ∫ cot2 x cot x dx

= ∫ cot3 x cosec2 x dx - ∫ (cosec2 x - 1) cot x dx

= ∫ cot3 x cosec2 x dx - ∫ cosec2 x cot x dx + ∫ cot x dx

Put, cot x = t

- cosec2 x dx = dt

= ∫- t3 dt - ∫(- t) dt + ∫ cot x dx

= - ∫ t3 dt + ∫ t dt + ∫ cot x dx

Integrate the above equation then, we get

= -t4 /4 + t2 /2 + log|sinx| + c

= -(cotx)4 /4 + (cotx)2 /2 + log|sinx| + c

Hence, I =-cot4 x /4 + cot2 x /2 + log|sinx| + c

Question 12. Integrate ∫ cot6 x dx

**Solution:

Let, I =∫ cot6 x dx

We can modify the above equation as below,

= ∫ cot2 x cot4 x dx

= ∫ (cosec2 x - 1) cot4 x dx

= ∫ (cosec2 x cot4 x - cot4 x) dx

= ∫ cosec2 x cot4 x - cot4 x dx

= ∫cosec2 x cot4 x dx - ∫ cot2 x cot2 x dx

= ∫cosec2 x cot4 x dx - ∫ (cosec2 x - 1) cot2 x dx

= ∫cosec2 x cot4 x dx - ∫ cosec2 x cot2 x + cot2 x dx

= ∫cosec2 x cot4 x dx - ∫ cosec2 x cot2 x dx+ ∫(cosec2 x -1) dx

= ∫cosec2 x cot4 x dx - ∫ cosec2 x cot2 x dx+ ∫cosec2 x dx - ∫1dx

Put, cot x = t

- cosec2 x dx = dt

= ∫ t4 (-dt) - ∫ t2 (-dt) + ∫ cosec2 x dx - ∫ dx

= -∫ t4 dt + ∫ t2 dt + ∫ cosec2 x dx - ∫ dx

Integrate the above equation then, we get

= -t5/5 + t3/3 - cot x - x + c

= -(cotx)5/5 + (cotx)3/3 - cot x - x + c

Hence, I = -cot5 x/5 + cot3 x /3 - cot x - x + c

Practice Questions:

1. ∫ (x² + 1) / (x³ - x) dx

2. ∫ (x² - 2x + 3) / (x² - 1) dx

3. ∫ (2x + 1) / (x² - x - 2) dx

4. ∫ (x² + x + 1) / (x³ + 1) dx

5. ∫ (3x - 2) / (x² + 2x + 1) dx

6. ∫ (x³ + 1) / (x⁴ - 1) dx

7. ∫ (2x² + 3x - 1) / (x³ - x) dx

8. ∫ (x² + 2) / (x³ - 3x + 2) dx

9. ∫ (x + 1) / (x³ - 3x² + 3x - 1) dx

10. ∫ (2x² - x + 3) / (x³ + x² - x - 1) dx

Summary

Exercise 19.11 in RD Sharma's Class 12 Chapter on Indefinite Integrals focuses on integrating rational functions using partial fraction decomposition. The key techniques used in this exercise include:

1. Factoring the denominator

2. Partial fraction decomposition

3. Integration of simpler rational functions

4. Recognition of standard integral forms

The problems typically involve breaking down complex rational functions into simpler fractions, then integrating each part separately and combining the results.

What is partial fraction decomposition?

It's a method of breaking down a complex fraction into a sum of simpler fractions, making it easier to integrate.

When should I use partial fraction decomposition?

Use it when integrating rational functions where the denominator can be factored and its degree is greater than the numerator's degree.

What are the steps for partial fraction decomposition?

Factor the denominator, 2) Set up the partial fractions, 3) Find the coefficients, 4) Write the original fraction as a sum of partial fractions.

How do I handle repeated linear factors in the denominator?

For a factor (x - a)ⁿ, you'll need terms of the form A/(x-a), B/(x-a)², ..., up to K/(x-a)ⁿ.