Question

Integrate dx/sin3xtan3x

Answer 1

$\displaystyle\int\frac{\mathrm dx}{\sin3x\tan3x}=\int\frac{\cos3x}{\sin^23x}$

Set $u=\sin3x$, so that $\mathrm du=3\cos3x\,\mathrm dx$. Then the integral is

$\displaystyle\frac13\int\frac{\mathrm du}{u^2}=-\frac1{3u}+C=-\frac1{3\sin3x}+C=-\frac13\csc3x+C$

Or, if you meant exponents in place of coefficients,

$\displaystyle\int\frac{\mathrm dx}{\sin^3x\tan^3x}=\int\csc^3x\cot^3x\,\mathrm dx=\int\csc^2x\cot^2x\csc x\cot x\,\mathrm dx$

Let $u=\csc x$, so that $\mathrm du=-\csc x\cot x\,\mathrm dx$, and use the fact that $\csc^2x=\cot^2x+1$ to get

$\displaystyle-\int u^2(u^2-1)\,\mathrm du=\int (u^2-u^4)\,\mathrm du=\frac13u^3-\frac15u^5+C$
$=\dfrac13\csc^3x-\dfrac15\csc^5x+C$