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