Question

 

 

Evaluate : `∫_0^(π/2)(sin^2 x)/(sinx+cosx)dx`

 

 

Answer 1

 

 

`Let I=∫_0^(π/2)(sin^2x)sinx+cosxdx`

`=>I=int_0^(pi/2)(sin^2(pi/2-x))/(sin(pi/2-x)+cos(pi/2-x)) dx   [Using :∫_0^a f(x)dx=∫_0^a f(a−x)dx]`

`=>I int_0^(pi/2) cos^2x/(cosx+sinx)dx  ..(ii)`

Adding (i) and (ii), we get

`2I=∫_0^(pi/2) (sin^2x)/(sinx+cosx)+cos^2x/(sinx+cosx)dx=∫0_^(pi/2)1/(sinx+cosx)dx`

`2I=int_0^(pi/2) 1/((2 tan (x/2))/(1+tan^2(x/2))+(1-tan^2(x/2))/(1+tan^2(x/2)))dx`

`=>2I=int_0^(pi/2)(1+tan^2(x/2))/(2tan(x/2)+1-tan^2(x/2))dx`

`=>2I=int_0^(pi/2)(sec^2(x/2))/(2 tan(x/2)+1-tan^2(x/2))dx`

`Let tan(x/2)=t. Then, d (tan(x/2))=dt⇒sec^2(x/2)⋅1/2dx=dt⇒sec^2 (x/2)dx=2 dt`

`Also x=0⇒t=tan 0=0 and x=(π/2)⇒t=tan(π/4)=1`

`∴ 2I=∫_0^1 (2dt)/(2t+1−t^2)=2∫_0^1 1/((sqrt2)^2−(t−1)^2)dt`

`=> 2I=2 xx1/(2sqrt2)[log|(sqrt2+t-1)/(sqrt2-t+1)|]_0^1`

`=>2I=1/sqrt2 {log(sqrt2/sqrt2)-log((sqrt2-1)/(sqrt2+1))}=1/sqrt2 {0-log((sqrt2-1)/(sqrt2+1))}`

`=>2I=-1/sqrt2 log {(sqrt2-1)^2/((sqrt2+1)(sqrt2-1))}=-1/sqrt2 log(sqrt2 -1)^2=-2/sqrt2 log (sqrt2-1)`

`=>I=-1/sqrt2 log (sqrt2-1)`