On page 263 of *A Course of Modern Analysis* by Whittaker & Watson, one is asked to show that

$$int_0^{frac{pi }{2}} cos

^{p+q-2}(theta ) cos

((p-q) theta ) , dtheta

=frac{pi }{(p+q-1)

2^{p+q-1} B(p,q)}$$

(*p*+*q*<1) where *B*(*p*,*q*) is the Beta function.The suggested approach is to evaluate the contour integral of $z^{p-q-1}

left(z+frac{1}{z}right)^

{p+q-2}$ around a closed contour consisting of the straight line beween *z*=-*i*,*i* and the semicircle |z|=1 in the right half plane (indented at the branch point *z*=0,*i*,-*i*). I am having trouble evaluating the contribution from the two segments along the imaginary axis.They can be combined into the single integral

$$-i int_{-1}^1 (i y)^{p-q-1}

left(i

y-frac{i}{y}right)^{p+q-2

} , dy$$

Mathematica can evaluate this integral and returns the value

$$-frac{i sin (pi q) Gamma

(1-q) Gamma

(p+q-1)}{Gamma (p)}$$

which can easily be shown to be equivalent to

$$-frac{i pi }{(p+q-1) B(p,q)}$$

How do I derive this result ?