Integrate the function

Dear student,

\int \frac{1}{{sinxcos}^{3} x} dx 1 = \sin^{2} x + \cos^{2} x \int \frac{\sin^{2} x + \cos^{2} x}{sinx . \cos^{3} x} \int \frac{\sin^{2} x}{sinx . \cos^{3} x} dx + \int \frac{\cos^{2} x}{sinx . \cos^{3} x} dx \int \frac{sinx}{\cos^{3} x} dx + \int \frac{1}{sinx . cosx} dx \int tanx . \sec^{2} xdx + \int \frac{1}{sinx . cosx} dx (\frac{sinx}{cosx} = tanx, \frac{1}{cosx} = secx) taking first integration \int tanx . \sec^{2} xdx let tanx = t \sec^{2} xdx = dt \int tdt = \frac{t^{2}}{2} + c = \frac{{(tanx)}^{2}}{2} + C taking second integration \int \frac{1}{sinx . cosx} dx multiplying and dividing by 2 \int \frac{2}{2 sinx . cosx} dx \int \frac{2}{\sin 2 x} dx = \int 2 cosec 2 xdx = \frac{2}{2} \log [cosec 2 x - \cot 2 x] = \log [cosec 2 x - \cot 2 x] + C Hence I = \frac{{(tanx)}^{2}}{2} + \log [cosec 2 x - \cot 2 x] + C

Regards