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Evaluate: π∫02π11+esinxdx

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प्रश्न

Evaluate: `int_0^(2π) (1)/(1 + e^(sin x)`dx

योग
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उत्तर

Let I = `int_0^(2π) (1)/(1 + e^(sin x)`dx   ...(i)

Applying property,

`int_0^af(x)dx = int_0^af(a-x)dx,` we get

I = `int_0^(2pi) dx/(1+e^(sin(2pi-x)))`

= `int_0^(2pi)dx/(1+e^(-sinx))`

= `int_0^(2pi)dx/(1+1/e^(sinx))`

= `int_0^(2pi)(e^(sinx)dx)/(e^(sinx)+1)`   ...(ii)

On adding equations (i) and (ii), we get

2I = `int_0^(2pi)dx/(1+e^(sinx))+int_0^(2pi)(e^(sinx)dx)/(1+e^(sinx))`

= `int_0^(2pi)((1+e^(sinx))/(1+e^(sinx)))dx`

= `int_0^(2pi)1.dx`

⇒ 2I = `[x]_0^(2pi)`

⇒ 2I = [2π]

⇒ I = π

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2021-2022 (March) Term 2 - Outside Delhi Set 1

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