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

### Question 1. Integrate ∫ tan^{3} x sec^{2} x dx

**Solution:**

Let, I = ∫ tan

^{3}x sec^{2}x dx ………. (i)Put,

tan x = t ………….. (ii)

d(tanx) = dt

sec

^{2}x dx = dt ………….. (iiI)Put equ (ii) and (iii) in equ (i), we get

= ∫ t

^{3}dtIntegrate the above equation, we get

= t

^{3+1}/3+1 + c= t

^{4}/ 4 + cPut the value of t from equ (ii), we get

= (tan x)

^{4}/ 4 + cHence, I = tan

^{4}x / 4 + c

### Question 2. Integrate ∫ tanx sec^{4} x dx

**Solution:**

Let, I = ∫ tan x sec

^{4}x dxWe can write the above equation as below,

= ∫ tanx sec

^{2}x sec^{2 }x dx= ∫ tanx (1+tan

^{2}x) sec^{2}x dx= ∫ (tanx + tan

^{3}x) sec^{2}x dx ………. (i)Substituting,

tanx = t ……….. (ii)

d(tanx) = dt

sec

^{2}dx = dt ……….. (iii)put equ (ii) and (iii) in equ (i), we get

= ∫ (t + t

^{3}) dtIntegrate the above equation, we get

= t

^{2}/2 + t^{3+1}/3+1 + c= t

^{2}/2 + t^{4}/4 + cput the value of t in the above equ then, we get

= (tanx)

^{2}/2 + (tanx)^{4}/ 4 + cHence, I = tan

^{2}x/2 + tan^{4}x / 4 + c

### Question 3. Integrate ∫ tan^{5} x sec^{4} x dx

**Solution:**

Let, I = ∫ tan

^{5}x sec^{4}x dxOn solving the above equation,

= ∫ tan

^{4 }x sec^{2}x sec^{2}xdx= ∫ tan

^{4}x (1+ tan^{2}x) sec^{2}xdx= ∫ (tan

^{5}x + tan^{7}x) sec^{2}xdx ………… (i)Put, tan x = t ………… (ii)

sec

^{2}x dx = dt ……….. (iii)Put equ (ii) and (iii) in equ (i), we get

= ∫ t

^{5}+ t^{7}dtIntegrate the above equation, we get

= t

^{6}/6 + t^{8}/8 + c= (tan x)

^{6}/6 + (tan x)^{8}/8 + cHence, I = tan

^{6}x/6 + tan^{8}x/8 + c

### Question 4. Integrate ∫ sec^{6} x tanx dx

**Solution:**

Let, I = ∫ sec

^{6}x tanx dxOn solving the above equation,

= ∫ sec

^{5}x (sec x tanx) dxPut, sec x = t

sec x tan x dx = dt

= ∫ t

^{5}dtIntegrate the above equation, we get

= t

^{6 }/6 + c= (sec x)

^{6}+ cHence, I = sec

^{6}x/6 + c

### Question 5. Integrate ∫ tan^{5} x dx

**Solution:**

Let, I = ∫ tan

^{5}x dxWe can modify the above equation as below,

= ∫ tan

^{2}x tan^{3}x dx= ∫ (sec

^{2}x – 1) tan^{3}x dx= ∫ sec

^{2}x tan^{3}x – tan^{3}x dx= ∫ sec

^{2}x tan^{3}x dx – ∫tan^{3}x dx= ∫ sec

^{2}x tan^{3}x dx – (∫(sec^{2}x – 1) tan x dx)= ∫ sec

^{2}x tan^{3}x dx – ∫ sec^{2}x tan x dx + ∫ tan x dxPut, tan x = t

sec

^{2}x dx = dt= ∫ t

^{3}dt – ∫ t dt + ∫ tan x dxNow, Integrate the above equation

= t

^{4}/4 – t^{2}/2 + log |sec x| + c= (tanx)

^{4}/4 – (tanx)^{2}/2 + log |sec x| + cHence, I = tan

^{4}x/4 – tan^{2}x /2 + log |sec x| + c

### Question 6. Integrate ∫ √tanx.sec^{4} x dx

**Solution:**

Let, I = ∫ √tanx.sec

^{4}x dx= ∫ √tanx.sec

^{2}x sec^{2}x dx= ∫ √tanx.(1 + tan

^{2}x) sec^{2}x dx= ∫ tan

^{1/2}x.(1 + tan^{2}x) sec^{2}x dx= ∫ (tan

^{1/2}x + tan^{5/2}x) sec^{2}x dxPut, tan x = t

sec

^{2}x dx = dt= ∫ (t

^{1/2}+ t^{5/2}) dtNow, 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}+ cHence, I = (2/3) tan

^{3/2}x + (2/7) tan^{7/2}x + c

### Question 7. Integrate ∫ sec^{4} 2x dx

**Solution:**

Let, I = ∫ sec

^{4}2x dx= ∫ sec

^{2}2x . sec^{2}2x dx= ∫(1 + tan

^{2}2x) sec^{2}2x dx= ∫ sec

^{2}2x + sec^{2}2x. tan^{2}2x dx= ∫ sec

^{2}2x. tan^{2}2x dx + ∫ sec^{2}2x dxPut, tan 2x = t

sec

^{2}2x dx = dt/2= ∫ t

^{2}dt/2 + ∫ sec^{2}2x dxIntegrate the above equation then, we get

= 1/2 × t

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

^{3}+ 1/2 tan 2x + cHence, I = = 1/2 tan2x + 1/6 tan

^{3}2x + c

### Question 8. Integrate ∫ cosec^{4} 3x dx

**Solution:**

Let, I = ∫ cosec

^{4}3x dx= ∫ cosec

^{2}3x cosec^{2}3x dx= ∫ (1+ cot

^{2}3x) cosec^{2}3x dx= ∫ (cosec

^{2}3x + cosec^{2}3x cot^{2}3x) dx= ∫ cosec

^{2}3x dx +∫ cosec^{2}3x cot^{2}3x dxPut, cot 3x = t

cosec

^{2}3x dx = – dt/3= ∫ cosec

^{2}3x dx – ∫ t^{2}dt/3Integrate the above equation then, we get

= -cot 3x/3 – t

^{3}/9 + c= (-1/3) cot 3x – (cot 3x)

^{3}/9 + cHence, I = (-1/3) cot 3x – (1/9) cot

^{3}3x + c

### Question 9. Integrate ∫ cot^{n} x cosec^{2} x dx (n ≠ -1)

**Solution:**

Let, I = ∫ cot

^{n}x cosec^{2}x dx, (n≠ -1)Put, cot x = t

– cosec

^{2}x dx = dt= – ∫ t

^{n}dtIntegrate the above equation, we get

= – t

^{n+1}/(n+1) + c= – (cot x)

^{n+1}/ (n+1) + cHence, I = – cot

^{n+1}x/ (n+1) + c

### Question 10. Integrate ∫ cot^{5} x cosec^{4} x dx

**Solution:**

Let, I = ∫ cot

^{5}x cosec^{4}x dx= ∫ cot

^{5}x cosec^{2}x cosec^{2}x dx= ∫ cot

^{5}x (1+ cot^{2}x) cosec^{2}x dx= ∫ (cot

^{5}x + cot^{7}x) cosec^{2}x dxPut, cot x = t

– cosec

^{2}x dx = dt= – ∫ (t

^{5}+ t^{7}) dtIntegrate the above equation, we get

= – t

^{6}/6 – t^{8}/8 + c= -(cotx)

^{6}/6 – (cotx)^{8}/8 + cHence, I = -cot

^{6}x/6 – cot^{8}x/8 + c

### Question 11. Integrate ∫ cot^{5} x dx

**Solution:**

Let, I = ∫ cot

^{5}x dxWe can modify the above equation as below,

= ∫ cot

^{3}x cot^{2}x dx= ∫ cot

^{3}x (cosec^{2}x – 1) dx= ∫ cot

^{3}x cosec^{2}x – cot^{3}x dx= ∫ cot

^{3}x cosec^{2}x dx – ∫ cot^{2}x cot x dx= ∫ cot

^{3}x cosec^{2}x dx – ∫ (cosec^{2}x – 1) cot x dx= ∫ cot

^{3}x cosec^{2}x dx – ∫ cosec^{2}x cot x dx + ∫ cot x dxPut, cot x = t

– cosec

^{2}x dx = dt= ∫- t

^{3}dt – ∫(- t) dt + ∫ cot x dx= – ∫ t

^{3}dt + ∫ t dt + ∫ cot x dxIntegrate the above equation then, we get

= -t

^{4}/4 + t^{2}/2 + log|sinx| + c= -(cotx)

^{4}/4 + (cotx)^{2}/2 + log|sinx| + cHence, I =-cot

^{4}x /4 + cot^{2}x /2 + log|sinx| + c

### Question 12. Integrate ∫ cot^{6} x dx

**Solution:**

Let, I =∫ cot

^{6}x dxWe can modify the above equation as below,

= ∫ cot

^{2}x cot^{4}x dx= ∫ (cosec

^{2}x – 1) cot^{4}x dx= ∫ (cosec

^{2}x cot^{4}x – cot^{4}x) dx= ∫ cosec

^{2}x cot^{4}x – cot^{4}x dx= ∫cosec

^{2}x cot^{4}x dx – ∫ cot^{2}x cot^{2}x dx= ∫cosec

^{2}x cot^{4}x dx – ∫ (cosec^{2}x – 1) cot^{2}x dx= ∫cosec

^{2}x cot^{4}x dx – ∫ cosec^{2}x cot^{2}x + cot^{2}x dx= ∫cosec

^{2}x cot^{4}x dx – ∫ cosec^{2}x cot^{2}x dx+ ∫(cosec^{2}x -1) dx= ∫cosec

^{2}x cot^{4}x dx – ∫ cosec^{2}x cot^{2}x dx+ ∫cosec^{2}x dx – ∫1dxPut, cot x = t

– cosec

^{2}x dx = dt= ∫ t

^{4}(-dt) – ∫ t^{2}(-dt) + ∫ cosec^{2}x dx – ∫ dx= -∫ t

^{4}dt + ∫ t^{2}dt + ∫ cosec^{2}x dx – ∫ dxIntegrate the above equation then, we get

= -t

^{5}/5 + t^{3}/3 – cot x – x + c= -(cotx)

^{5}/5 + (cotx)^{3}/3 – cot x – x + cHence, I = -cot

^{5}x/5 + cot^{3}x /3 – cot x – x + c

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