Question

Find integration of 2 cos x upon 1 minus sin x into 1 + sin square x dx

Answer 1

Answer:

As, we have to find the integration of,

$\int {\frac{2cosx\times(1+sin^2x)}{1-sinx}} \, dx\\\\ {\text{put}, sin x=t, {\text{then}} cosx dx=dt$

$=\int \frac{2(1+t^2)}{1-t} \, dt \\\\= \int {\frac{2}{1-t} \, dt +2 \int {\frac{t^2}{1-t}} \, dt \\\\ =\int {\frac{2}{1-t} \, dt -2 \int {\frac{1-t^2-1}{1-t}} \, dt$

$=\int {\frac{2}{1-t} \, dt -\int {\frac{2(1-t^2)}{1-t} \, dt+\int {\frac{2}{1-t} \, dt\\\\=\int {\frac{2}{1-t} \, dt - \int {2(1+t)} \, dt+\int \frac{2}{1-t} \, dt$

=4 log|1-t|-2 t + $\frac{2t^2}{2}+C$

$=4log|1-sinx | -2sin x +sin^2x+C$