Solve $$\int_{\pi/3}^\pi \cos\frac{1}{\sin(\csc x)}\cdot\cos(\csc x)\cdot\frac{\cos x}{\sin^2x\cdot\sin^2(\csc x)}\ dx$$
How to start. First simplify trigonometrically then differentiate? then it is more of a trigonometry question Or substitute something to $t$ and find out what is $dt/dx$ and then go further?
Something as nasty as this is only going to be solvable if it's the derivative of a nasty composite. My first guess is it's the derivative of $\sin{\csc{\csc{x}}}$ (I took the part with most compositions and wrote down the integral of the $\cos$ part), so let's differentiate and see what happens: \begin{align} \frac{d}{dx} \sin{\csc{\csc{x}}} &= (\cos{\csc{\csc{x}}}) \frac{d}{dx} \csc{\csc{x}} \\ &= -(\cos{\csc{\csc{x}}}) (\csc{\csc{x}}\cot{\csc{x}})\frac{d}{dx} \csc{x} \\ &= +(\cos{\csc{\csc{x}}}) \frac{\cos{\csc{x}}}{\sin^2{\csc{x}}} (\csc{x}\cot{x}) \\ &= (\cos{\csc{\csc{x}}}) \frac{\cos{\csc{x}}}{\sin^2{\csc{x}}} \frac{\cos{x}}{\sin^2{x}}, \end{align} which of course is exactly the integrand. Hence the integral is $$ \int_{\pi/3}^{\pi} \frac{d}{dx} \sin{\csc{\csc{x}}} \, dx, $$ from which it's easy to finish. (Note that although the integrand behaves quite badly at $x=\pi/2$, we can still find the continuous antiderivative we needed for the Fundamental Theorem of Calculus to work.)
Hint put $t=\sin (\csc (x)) $ thus $dt=-\cos (csc (x))\csc(x)\cot (x)$ . you will see that a large part of the expression becomes dt and then its solvable
