The question is: how can I prove that:
$$\int_{0}^{\pi} \sin^n\theta\ d\theta = \frac{\Gamma\big(\frac{1}{2}\big) \Gamma\big(\frac{1}{2} + \frac{1}{2}n\big)}{\Gamma\big(1 + \frac{1}{2}n\big)}$$
The question is: how can I prove that:
$$\int_{0}^{\pi} \sin^n\theta\ d\theta = \frac{\Gamma\big(\frac{1}{2}\big) \Gamma\big(\frac{1}{2} + \frac{1}{2}n\big)}{\Gamma\big(1 + \frac{1}{2}n\big)}$$
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The simple way, IMHO, is to check that identity holds for $n=0,1$ and both the LHS and the RHS satisfy the same recurrence relation; the LHS in virtue of integration by parts, the RHS in virtue of $\Gamma(z+1)=z\cdot\Gamma(z)$.