# Show that $\int_{0}^{\frac \pi2} f(x) \operatorname{d}x = \pi/4$

$f(x) = \dfrac{\sin^\alpha x \operatorname{d}x}{\sin^\alpha x + \cos^\alpha x}$

Well, if $\alpha$ = 0,

$$\int_{0}^{\frac \pi2} \frac 12 dx = \frac 12 \frac \pi2 = \frac \pi4$$

But I don't know what other cases I need to do ($\alpha > 0$ and $\alpha < 0$ ?)

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We don't need to treat $\alpha=0$ separately.

$$\text{We know, }\int_a^b f(x)dx=\int_a^b f(a+b-x)dx$$

$$\text{So, }\int_0^\frac\pi2\frac{\sin^{\alpha}x}{\sin^{\alpha}x+\cos^{\alpha}x}dx$$

$$=\int_0^\frac\pi2\frac{\sin^{\alpha}\left(\frac\pi2+0-x\right)}{\sin^{\alpha}\left(\frac\pi2+0-x\right)+\cos^{\alpha}\left(\frac\pi2+0-x\right)}dx$$

$$=\int_0^\frac\pi2\frac{\cos^{\alpha}x}{\sin^{\alpha}x+\cos^{\alpha}x}dx=I\text{(say)}$$

$$\text{So, } I+I=\int_0^\frac\pi2\frac{\sin^{\alpha}x}{\sin^{\alpha}x+\cos^{\alpha}x}dx+\int_0^\frac\pi2\frac{\cos^{\alpha}x}{\sin^{\alpha}x+\cos^{\alpha}x}dx=\int_0^\frac\pi2dx$$ assuming $\sin^{\alpha}x+\cos^{\alpha}x\ne0$ as it would make the given integral undefined.

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@David. $2I=\frac\pi2$ right? – lab bhattacharjee Mar 25 '13 at 16:40
Yes, I didn't see before. – David Mar 25 '13 at 16:42