Solveeit Logo

Question

Mathematics Question on integral

dx1cosxsinx\int\frac{dx}{1-cos x -sin x} is equal to

A

log1+cotx2+Clog \left|1+cot \frac{x}{2}\right| +C

B

log1tanx2+Clog \left|1-tan \frac{x}{2}\right| +C

C

log1cotx2+Clog \left|1-cot \frac{x}{2}\right| +C

D

log1+tanx2+Clog \left|1+tan \frac{x}{2}\right| +C

Answer

log1cotx2+Clog \left|1-cot \frac{x}{2}\right| +C

Explanation

Solution

LetI=dx1cosxsinx I = \int \frac{dx}{1-cos x -sin x} put cosx=1tan2x21+tan2x2andsinx=2tanx21+tan2x2cos x = \frac{1-tan^{2} \frac{x}{2}}{1+tan^{2} \frac{x}{2}} and sin x= \frac{2tan \frac{x}{2}}{1+tan ^{2 } \frac{x}{2}} I=dx1(1tan2x21+tan2x2)2tanx21+tan2x2\therefore I= \int\frac{dx}{1-\left(\frac{1-tan^{2} \frac{x}{2}}{1+tan^{2} \frac{x}{2}}\right)- \frac{2tan \frac{x}{2}}{1+tan ^{ 2} \frac{x}{2}}} =sec2x2dx2tan2x22tanx2=12sec2x2dxtan2x2tanx2= \int \frac{sec ^{2} \frac{x}{2} dx }{2tan ^{2} \frac{x}{2}-2tan \frac{x}{2}} = \int\frac{\frac{1}{2}sec^{2} \frac{x}{2} dx }{tan^{2} \frac{x}{2} -tan \frac{x}{2}} Put tanx2=t12sec2x2dx=dttan \frac{x}{2}=t \Rightarrow \frac{1}{2} sec^{2} \frac{x}{2 }dx = dt I=dtt2t=[1t11t]+C\therefore I = \int \frac{dt}{t^{2} -t} = \left[\frac{1}{t-1}-\frac{1}{t}\right]\int+C =log(t1)logt+C=logt1t+C=log\left(t-1\right) -logt + C = log\left|\frac{t-1}{t} +C\right| =log1cotx2+C=log \left|1-cot \frac{x}{2}\right| +C