Question

Integrate dx/sin3xtan3x

Answer 1

`\displaystyle\int(\mathrm dx)/(\sin3x\tan3x)=\int(\cos3x)/(\sin^23x)`

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


`\displaystyle\frac13\int(\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(\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`

`=\frac13\csc^3x-\frac15\csc^5x+C`