-0-2 cos x-1-sin xdx-

Put 1+ sin x =t Then ∫_0^π/2cos x/1+sin xdx = [ log |1+sin x| ]^π/2_0 =log 2 ...

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Property 1

∫0π/2 cos x/1...

Question

$\int_{0}^{\frac{π}{2}} \frac{c o s x}{1 + s i n x} d x =$

log 2

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log e

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$\frac{1}{2} l o g 3$

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Solution

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\int_{0}^{\frac{π}{2}} \frac{c o x x - s i n x}{1 + s i n x c o s x} d x =

\int_{0}^{\frac{π}{2}} \frac{c o s x}{(1 + s i n x) (2 + s i n x)} d x =

\int_{0}^{\frac{π}{2}} \frac{c o s x}{(1 + s i n x) (2 + s i n x)} d x =

$\int \frac{s i n^{- 1} \sqrt{x} - c o s^{- 1} \sqrt{x}}{s i n^{- 1} \sqrt{x} + c o s^{- 1} \sqrt{x}} d x .$

I = \int c o s^{- 1} \sqrt{x} d x

J = \int \frac{s i n^{- 1} \sqrt{x} - c o s^{- 1} \sqrt{x}}{s i n^{- 1} \sqrt{x} + c o s^{- 1} \sqrt{x}} d x

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