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tatyana61 [14]
3 years ago
15

Integrate dx/sin3xtan3x

Mathematics
1 answer:
disa [49]3 years ago
8 0
\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
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