Evaluate the following integral: `int_(1//3)^1((x-x^3)^(1//3))/(x^4)dx`

Explanation

Let `I=int_(1//3)^1((x-x^3)^(1//3))/(x^4)dx`

Put `x=sinθ
⇒dx=cosθdθ`

When `x=1/3​,θ=sin^(−1)(1/3​)` and when `x=1,θ=π/2`,

​`I=int_(sin^(-1).(1//3))^(pi/2)((sinθ-sin^3theta)^(1//3))/(sin^4theta)costheta d theta`

​`I=int_(sin^(-1).(1//3))^(pi/2) ((sinθ)^(1/3)​(1−sin^2θ)^(1/3)​​)/sin^4theta cosθdθ`

=`int_(sin^(-1).(1//3))^(pi/2) ((sinθ)^(1/3​)(cosθ)^(2/3))/(sin^2θsin^2θ)cosθdθ`

=`int_(sin^(-1).(1//3))^(pi/2) (cotθ)^(5/3)​cosec^2θdθ`

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