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My approach is to use the properties of the beta function. But the limits of integration are from zero to $\frac{\pi}{4}$. We know that

$$ B(m,n) = 2\int_{0}^{\pi/2}\sin^{2m-1}\theta \cos^{2n-1}\theta d\theta = \dfrac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)} $$

Thanks for any help.

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  • $\begingroup$ $$\sin^{n+1}t\cos^{n-1}t=\dfrac{\tan^{n+1}t\sec^2t}{\sec^{2n+2}t}$$ $\endgroup$ – lab bhattacharjee Jul 25 '18 at 14:32
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Hint: Use substitution $\theta=\dfrac{u}{2}$ then continue with $$\dfrac{1}{2^{n+1}}\int_0^{\pi/2}\sin^{n-1}u(1-\cos u)du=\dfrac{\beta(\frac{n}{2},\frac12)-\beta(\frac{n}{2},1)}{2^{n+2}}$$

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  • $\begingroup$ The so-called Weierstrass substitution works miracles with trigonometric integrals. $\endgroup$ – Pedro Tamaroff Jul 26 '18 at 2:02

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