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Evaluate : ∫0π2cosx(1+sinx)(2+sinx)⋅dx - Mathematics and Statistics

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Sum

Evaluate : `int_0^(pi/2) cosx/((1 + sinx)(2 + sin x))*dx`

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Solution

Let I = `int_0^(pi/2) cosx/((1 + sinx)(2 + sin x))*dx`
Put sin x = t
∴ cos x·dx = dt

When x = `pi/(2), t = sin  pi/(2)` = 1

When x = 0, t = sin 0 = 0

∴ I = `int_0^1 dt/((1 + t)(2 + t)`

= `int_0^1((2 + t) - (1 + t))/((1 + t)(2 + t))*dt`

= `int_0^1[1/(1 + t) - 1/(2 + t)]*dt`

= `int_0^1 1/(1 + t)*dt - int_0^1 1/(2 + t)*dt`

= `[log |1 + t|]_0^1 - [log|2 + t|]_0^1`
= [log(1 + 1) – log(1 + 0)] – [log(2 + 1) – log(2 + 0)]
= log 2 – log 3 + log 2    ...[∵ log 1 = 0]

= `log ((2 xx 2)/3)`

= `log(4/3)`.

Concept: Fundamental Theorem of Integral Calculus
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