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

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.

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

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• What work have you done so far? – Arjun Dhiman Jul 20 '15 at 20:25
• What work have you done so far? – tired Jul 20 '15 at 20:45
• $\frac{\pi}{4}$ is the right result – Dr. Sonnhard Graubner Jul 20 '15 at 21:06

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.