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Question
Evaluate: `int_0^(pi/4) cosx/(4 - sin^2 x) "d"x`
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Solution
Let I = `int_0^(pi/4) cosx/(4 - sin^2 x) "d"x`
Put sin x = t
∴ cos x dx = dt
When x = 0, t = 0 and when x = `pi/4`,t `1/sqrt(2)`
∴ I = `int_0^(1/sqrt(2)) "dt"/(4 - "t"^2)`
= `int_0^(1/sqrt(2)) "dt"/(2^2 - "t"^2)`
= `[1/(2 xx 2) log|(2 + "t")/(2 - "t")|]_0^(1/sqrt(2))`
= `1/4[log|(2 + 1/sqrt(2))/(2 - 1/sqrt(2))| - log 1]`
= `1/4[log|(2sqrt(2) + 1)/(2sqrt(2) - 1)| - 0]`
∴ I = `1/4 log((2sqrt(2) + 1)/(2sqrt(2) - 1))`
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