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Question
Show that `int_0^(pi/2) (sin^2x)/(sinx + cosx) = 1/sqrt(2) log (sqrt(2) + 1)`
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Solution
We have I = `int_0^(pi/2) (sin^2x)/(sinx + cosx) "d"x`
= `int_0^(pi/2) (sin^2(pi/2 - x))/(sin(pi/2 - x) + cos(pi/2 - x)) "d"x` ....(By P4)
⇒ I = `int_0^(pi/2) (cos^2x)/(sinx + cosx) "d"x`
Thus, we get 2I = `1/sqrt(2) int_0^(pi/2) ("d"x)/(cos(x - pi/4))`
= `1/sqrt(2) int_0^(pi/2) sec(x - pi/2) "d"x`
= `1/sqrt(2) [log(sec(x - pi/4) + tan(x - pi/4))]_0^(pi/2)`
= `1/sqrt(2)[log(sec pi/4 + tan pi/4) - log sec(- pi/4) + tan(- pi/4)]`
= `1/sqrt(2) [log(sqrt(2) + 1) - log(sqrt(2) - 1)]`
= `1/sqrt(2) log|(sqrt(2) + 1)/(sqrt(2) - 1)|`
= `1/sqrt(2) log((sqrt(2) - 1)^2/1)`
= `2/sqrt(2) log(sqrt(2) + 1)`
Hence I = `1/sqrt(2) log(sqrt(2) + 1)`.
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