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Question
Evaluate the following : `int_(pi/4)^(pi/2) (cos theta)/[cos theta/2 + sin theta/2]^3*d theta`
Solution
Let I = `int_(pi/4)^(pi/2) (cos theta)/[cos theta/2 + sin theta/2]^3*d theta`
= `int_(pi/4)^(pi/2) (cos^2 theta/2 - sin^2 theta/2)/[cos theta/2 + sin theta/2]^3*d theta`
= `int_(pi/4)^(pi/2)((cos theta/2 - sin theta/2)(cos theta/2 + sin theta/2))/[cos theta/2 + sin theta/2]^3*d theta`
= `int_(pi/4)^(pi/2) (cos theta/2 - sin theta/2)/[cos theta/2 + sin theta/2]^2*d theta`
Put `cos theta/2 - sin theta/2` = t
∴ `(-1/2 sin theta/2 +1/2 cos theta/2)*d theta` = dt
∴ `(cos theta/2 - sin theta/2)*d theta = 2*dt`
When θ = `pi/(4), t = cos pi/(8) + sin pi/(8)`
When θ = `pi/(2), t = cos pi/(4) + sin pi/(4) = (1)/sqrt(2) + (1)/sqrt(2) = sqrt(2)`
∴ I = `int_(cos pi/8 + sin pi/8)^sqrt(2) (1)/t^2* 2dt`
= `2 int_(cos pi/8 + sin pi/8)^sqrt(2) t^-2*dt`
= `2[(t^-1)/-1]_(cos pi/8 + sin pi/8)^sqrt(2)`
= `[(-2)/t]_(cos pi/8 + sin pi/8)^sqrt(2)`
= `- (2)/sqrt(2) + (2)/(cos pi/8 + sin pi/8)`
= `(2)/(cos pi/8 + sin pi/8) - sqrt(2)`.
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