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Question Number 106583 by bemath last updated on 06/Aug/20

@ bemath @

\underset{0}{\int^{π / 2}} \frac{\sin x dx}{\sin x + \cos x} = ?

Answered by bobhans last updated on 06/Aug/20

let a = \underset{0}{\int^{\frac{π}{2}}} \frac{\sin x dx}{\sin x + \cos x}; ♭ = \underset{0}{\int^{\frac{π}{2}}} \frac{\cos x dx}{\sin x + \cos x}

(1) a + ♭ = \underset{0}{\int^{\frac{π}{2}}} dx = \frac{π}{2}

(2) a - ♭ = \underset{0}{\int^{\frac{π}{2}}} \frac{\sin x - \cos x}{\sin x + \cos x} dx = - \underset{1}{\int^{1}} \frac{du}{u} = 0

[with u = \sin x + \cos x]; a = ♭

therefore a = \underset{0}{\int^{\frac{π}{2}}} \frac{\sin x dx}{\sin x + \cos x} = \frac{1}{2} \times \frac{π}{2} = \frac{π}{4}

Commented by john santu last updated on 06/Aug/20

nice & cooll . .

Commented by bemath last updated on 06/Aug/20

creative \dots \iddots

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