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