# Class 12 RD Sharma Solutions- Chapter 19 Indefinite Integrals – Exercise 19.11

Last Updated : 05 Mar, 2021

### Question 1. Integrate âˆ« tan3 x sec2 x dx

Solution:

Let, I = âˆ« 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

= âˆ« 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 âˆ« tanx sec4 x dx

Solution:

Let, I = âˆ« tan x sec4 x dx

We can write the above equation as below,

= âˆ« tanx sec2 x sec2 x dx

= âˆ« tanx (1+tan2 x) sec2 x dx

= âˆ« (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

= âˆ« (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 âˆ« tan5 x sec4 x dx

Solution:

Let, I = âˆ« tan5 x sec4 x dx

On solving the above equation,

= âˆ« tan4 x sec2 x sec2 xdx

= âˆ« tan4 x (1+ tan2 x) sec2 xdx

= âˆ« (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

= âˆ« 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 âˆ« sec6 x tanx dx

Solution:

Let, I = âˆ« sec6 x tanx dx

On solving the above equation,

= âˆ« sec5 x (sec x tanx) dx

Put, sec x = t

sec x tan x dx = dt

= âˆ« t5 dt

Integrate the above equation, we get

= t6 /6 + c

= (sec x)6 + c

Hence, I = sec6 x/6 + c

### Question 5. Integrate âˆ« tan5 x dx

Solution:

Let, I = âˆ« tan5 x dx

We can modify the above equation as below,

= âˆ« tan2 x tan3 x dx

= âˆ« (sec2 x – 1) tan3 x dx

= âˆ« sec2 x tan3 x – tan3 x dx

= âˆ« sec2 x tan3 x dx – âˆ«tan3 x dx

= âˆ« sec2 x tan3 x dx – (âˆ«(sec2 x – 1) tan x dx)

= âˆ« sec2 x tan3 x dx – âˆ« sec2 x tan x dx + âˆ« tan x dx

Put, tan x = t

sec2 x dx = dt

= âˆ« t3 dt – âˆ« t dt + âˆ« 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 âˆ« âˆštanx.sec4 x dx

Solution:

Let, I = âˆ« âˆštanx.sec4 x dx

= âˆ« âˆštanx.sec2 x sec2 x dx

= âˆ« âˆštanx.(1 + tan2 x) sec2 x dx

= âˆ« tan1/2 x.(1 + tan2 x) sec2 x dx

= âˆ« (tan1/2 x + tan5/2 x) sec2 x dx

Put, tan x = t

sec2 x dx = dt

= âˆ« (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 âˆ« sec4 2x dx

Solution:

Let, I = âˆ« sec4 2x dx

= âˆ« sec2 2x . sec2 2x dx

= âˆ«(1 + tan2 2x) sec2 2x dx

= âˆ« sec2 2x + sec2 2x. tan2 2x dx

= âˆ« sec2 2x. tan2 2x dx + âˆ« sec2 2x dx

Put, tan 2x = t

sec2 2x dx = dt/2

= âˆ« t2 dt/2 + âˆ« sec2 2x dx

Integrate the above equation then, we get

= 1/2 Ã— 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 âˆ« cosec4 3x dx

Solution:

Let, I = âˆ« cosec4 3x dx

= âˆ« cosec2 3x cosec2 3x dx

= âˆ« (1+ cot2 3x) cosec2 3x dx

= âˆ« (cosec2 3x + cosec2 3x cot2 3x) dx

= âˆ« cosec2 3x dx +âˆ« cosec2 3x cot2 3x dx

Put, cot 3x = t

cosec2 3x dx = – dt/3

= âˆ« cosec2 3x dx – âˆ« 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 âˆ« cotn x cosec2 x dx (n â‰  -1)

Solution:

Let, I = âˆ« cotn x cosec2 x dx, (nâ‰  -1)

Put, cot x = t

– cosec2 x dx = dt

= – âˆ« 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 âˆ« cot5 x cosec4 x dx

Solution:

Let, I = âˆ« cot5 x cosec4 x dx

= âˆ« cot5 x cosec2 x cosec2 x dx

= âˆ« cot5 x (1+ cot2 x) cosec2 x dx

= âˆ« (cot5 x + cot7 x) cosec2 x dx

Put, cot x = t

– cosec2 x dx = dt

= – âˆ« (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 âˆ« cot5 x dx

Solution:

Let, I = âˆ« cot5 x dx

We can modify the above equation as below,

= âˆ« cot3 x cot2 x dx

= âˆ« cot3 x (cosec2 x – 1) dx

= âˆ« cot3 x cosec2 x – cot3 x dx

= âˆ« cot3 x cosec2 x dx – âˆ« cot2 x cot x dx

= âˆ« cot3 x cosec2 x dx – âˆ« (cosec2 x – 1) cot x dx

= âˆ« cot3 x cosec2 x dx – âˆ« cosec2 x cot x dx + âˆ« cot x dx

Put, cot x = t

– cosec2 x dx = dt

= âˆ«- t3 dt – âˆ«(- t) dt + âˆ« cot x dx

= – âˆ« t3 dt + âˆ« t dt + âˆ« 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 âˆ« cot6 x dx

Solution:

Let, I =âˆ« cot6 x dx

We can modify the above equation as below,

= âˆ« cot2 x cot4 x dx

= âˆ« (cosec2 x – 1) cot4 x dx

= âˆ« (cosec2 x cot4 x – cot4 x) dx

= âˆ« cosec2 x cot4 x – cot4 x dx

= âˆ«cosec2 x cot4 x dx – âˆ« cot2 x cot2 x dx

= âˆ«cosec2 x cot4 x dx – âˆ« (cosec2 x – 1) cot2 x dx

= âˆ«cosec2 x cot4 x dx – âˆ« cosec2 x cot2 x + cot2 x dx

= âˆ«cosec2 x cot4 x dx – âˆ« cosec2 x cot2 x dx+ âˆ«(cosec2 x -1) dx

= âˆ«cosec2 x cot4 x dx – âˆ« cosec2 x cot2 x dx+ âˆ«cosec2 x dx – âˆ«1dx

Put, cot x = t

– cosec2 x dx = dt

= âˆ« t4 (-dt) – âˆ« t2 (-dt) + âˆ« cosec2 x dx – âˆ« dx

= -âˆ« t4 dt + âˆ« t2 dt + âˆ« cosec2 x dx – âˆ« 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

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