Question

Evaluate `int_0^(pi/4) (sinx + cosx)/(16 + 9sin2x) dx`

Answer 1

Let I = `int_0^(pi/4) (sinx + cosx)/(16 + 9sin2x) dx`

Put -cosx + sin x = t .....(1)

Then

(sin x + cos x) dx= dt

As x → 0, t → -1

Also `x = pi/4`, t → 0

Squaring (1) both sides we get

`cos^2x + sin^2x - 2cosxsinx  = t^2`

`=> 1 - sin2x = t^2`

`=> sin 2x =1 - t^2`

Substituting these values, we get

`I = int_(-1)^0  dt/(16+9(1-t^2))`

`= int_(-1)^0  dt/(25 - 9t^2)`

`= 1/9 int_(-1)^0 dt/((5/3)^2 - t^2)`

`= 1/9[1/(2a) log |(a+t)/(a-t)|]_(-1)^0`     where a = 5/3

`= 1/9 [3/(2(5)) ""  log |(5/3+t)/(5/3-t)|]_(-1)^0`

`= 1/9 [3/10 {log 1 - log  1/4}]^(-1)`

`= 3/90 (-log  1/4) = 1/30 log 4`