RD Sharma Class 12 Ex 19.11 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.11 Solutions Chapter 19 Indefinite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest Sharma Class 12 Ex 19.11 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter19
Exercise19.11
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 19.11 Solutions Chapter 19 Indefinite Integrals

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 secx 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,

= ∫ tanx 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

= t/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

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