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I would like to evaluate the following integral:

$$\int_0^{\pi/2} \frac {\sqrt[3]{(\sin^2x)}} {\sqrt[3]{(\sin^2x)}+{\sqrt[3]{(\cos^2x)}}}\,\mathrm{d}x$$

Give me show step by step solutions please.

Thank you very much.

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closed as off-topic by user147263, Mike Pierce, Micah, graydad, J. W. Perry Jul 21 '15 at 1:34

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  • 2
    $\begingroup$ What work have you done so far? $\endgroup$ – Arjun Dhiman Jul 20 '15 at 20:25
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    $\begingroup$ What work have you done so far? $\endgroup$ – tired Jul 20 '15 at 20:45
  • $\begingroup$ $\frac{\pi}{4}$ is the right result $\endgroup$ – Dr. Sonnhard Graubner Jul 20 '15 at 21:06
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You can make use of this! $$ I = \int_{a}^{b} f(x) \text{ d}x = \int_{a}^{b} f(a+b-x) \text{ d}x $$

Try to evaluate both of the above and add them together. You should get $2I$ as equal to something. Then divide by $2$ and you'll find out what $I$ equals.

A pictorial argument as to why this is true is provided here.

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