मराठी

Evaluate: π∫0π211+(tanx)23dx

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प्रश्न

Evaluate: `int_0^(π/2) 1/(1 + (tanx)^(2/3)) dx`

बेरीज
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उत्तर

Let I = `int_0^(π/2) 1/(1 + (tanx)^(2/3)) dx`  ...(i)

I = `int_0^(π/2) 1/(1 + [tan(π/2 - x)]^(2/3)) dx`  ...[Using property `int_0^a f(x)dx = int_0^a f(a - x)dx`]

I = `int_0^(π/2) 1/(1 + (cot x)^(2/3)) dx`

I = `int_0^(π/2) ((tanx)^(2/3))/((tanx)^(2/3) + 1) dx`

I = `int_0^(pi/2) ((tanx)^(2/3) + 1 - 1)/((tanx)^(2/3) + 1) dx`

I = `int_0^(π/2) (1 + (tanx)^(3/2))/(1 + (tanx)^(3/2)) dx - int_0^(π/2) 1/(1 + (tanx)^(3/2)) dx`

I = `int_0^(π/2) 1.dx - I`  ...[From equation (i)]

2I = `int_0^(π/2) 1.dx`

2I = `[x]_0^(π/2)`

2I = `π/2`

I = `π/4`

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

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