RD Sharma Class 12 Ex 19.11 Solutions Chapter 19 Indefinite Integrals

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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 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

= ∫ tan^1/2 x.(1 + tan² x) sec² x dx

= ∫ (tan^1/2 x + tan^5/2 x) sec² x dx

Put, tan x = t

sec² x dx = dt

= ∫ (t^1/2 + t^5/2) dt

Now, Integrate the above equation,

= t^3/2/(3/2) + t^7/2/(7/2) + c

= (2/3) t^3/2 + (2/7) t^7/2 + c

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

Hence, I = (2/3) tan^3/2 x + (2/7) tan^7/2 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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