Evaluate the definite integral $ \int_{\pi/6}^{\pi/2} \frac{\cos(x)}{\sin^{5/7}(x)}\, dx$ Evaluate the integral:
$\displaystyle \int_{\pi/6}^{\pi/2} \frac{\cos(x)}{\sin^{5/7}(x)}\, dx$
(using substitution)
Here's my attempt at solution:
u = $\sin^5(x)$
$du = 5\sin^4(x) \cdot \cos(x) \cdot dx$
$ \frac {1}{5\sin^4(x)} du = \cos(x) \cdot dx $
Also, lower and upper limits for integration will be different for new variable u
since $u = \sin^5(x)$
new lower limit is $\frac {1}{32}$
and upper limit is $1$
Making substitution:
$\displaystyle \int_{\frac{1}{32}}^{1} \frac{1}{u^\frac{1}{7}} \cdot \frac{1}{5\sin^4(x)} du$
... and I'm stuck, I dunno how to take an integral of $\frac{1}{5\sin^4(x)} $
Solution shouldn't come to this, there must be another way.

Hint: $\cos x$ is the derivative of $\sin x$, and use the chain rule, where (the supersrcipt $n$ is the power, and $f'(x)$ is the derivative of $f(x)$ with respect to $x$)
$$\frac{d}{dx}(f(x)^n)=nf(x)^{n-1}f'(x)$$
