∫f(sinx)cosxdx　の形は
sinx=t　と置換すれば
大抵はうまくいくと思うのですが・・・

∫[π/6→π/2]cos(x)dx/(sin(x))^(\(\frac{1}{2}\))
(sinx=tとおくと)

=∫[\(\frac{1}{2}\)→1](1/\(\sqrt{\quad}\)t)dt
=[2\(\sqrt{\quad}\)t](t=1,t=\(\frac{1}{2}\))
=2-\(\sqrt{\quad}\)2