Integrate the following w.r.t. x : 1sin2x+cosx - Mathematics and Statistics

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Sum

Integrate the following w.r.t. x : `(1)/(sin2x + cosx)`

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Solution

Let I = `int (1)/(sin2x + cosx)*dx`

= `int (1)/(cosx + 2sinx cosx)*dx`

= `int (cosxdx)/(cosx(1 + 2 sinx)`

= `int (cosx*dx)/((1-sin^2x)(1 + 2sinx)`

= `int (cos*dx)/((1 - sin^2x)(1 + 2 sinx)`

= `int (cosx*dx)/((1 + sin^2x)(1 - sin)(1 + 2sinx)`
Put cos x = t

∴ t = sin x

dt = cosxdx

∴ I = `int (dt)/((1+ t)(1 - t)(2t + 1)`

= `-int (dt)/((1 + t)(1 - t)(2t+1)`

Let `(1)/((1 + t)(1 - t)(2t + 1)) = "A"/(1 +  t) + "B"/(1 - t) + "C"/(2t+ 1)`

∴ I = A (1-t) (2t + 1) + B (1 + t) (2t + 1) +C (1+t)(1-t)
Puttingt = t= 1

1 = B(1 + 1) (2×1+1)

1 = B ×2×3

`B = 1/6`
Putting 1 – t = 0, i.e. t = – 1, we  get
1 = A(0)(– 1) + B(2)(– 1) + C(2)(0)
∴ B = `-(1)/(2)`
Putting 1 + 2t = 0, i.e. t = `-(1)/(2)`, we get

1 = A (1 + 1) (-2 + 1)

1 = A ×2 ×-1

A = `(-1)/2`

2t + 1= 0

t = `(-1/2)`

`1= C (1-1/2) (1 + 1/2)`

`1= C × 1/2×3/2`

`1 = (3C)/4`

`C = 4/3`

∴ `1/((1 - t)(1 + t)(1 + 2t)) = ((1/6))/(1 - t) + (((-1)/2))/(1 + t) + ((4/3))/(1 + 2t)`

∴ I = `int [((1/6))/(1 - t) + (((-1)/2))/(1 + t) + ((4/3))/(1 + 2t)]*dt`

= `(1)/(6) int (1)/(1 - t)*dt + 1/2 int 1/(1 + t)*dt - 4/3 int 1/(1 + 2t)*dt`

= `(1)/(6)*(log |1 - t|)/(-1) + 1/2log|1 + t| - 4/3*(log|1 + 2t|)/(2) + c`

= `(1)/(6)log|sinx + 1| + 1/2log|sinx - 1| - 2/3log|sinx 1 + 2| + c`

= `(1)/(6)log|1 - sinx| - (1)/(2)log|1 + sinx| + (2)/(3)log|1 + 2sinx| + c`.

  Is there an error in this question or solution?
Chapter 3: Indefinite Integration - Exercise 3.4 [Page 145]

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