Question

$\int \frac{dx}{1-cosx-sinx}$ is equal to

• A. $log\left|1+cot\frac{x}{2}\right|+C$
• B. $log\left|1-tan\frac{x}{2}\right|+C$
• C. $log\left|1-cot\frac{x}{2}\right|+C$
• D. $log\left|1+tan\frac{x}{2}\right|+C$

Answer 1

Answer: (C)

Let$I=\int \frac{dx}{1-cosx-sinx}$

put $cosx=\frac{1-ta{n}^2\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}andsinx=\frac{2tan\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}$

$\therefore I=\int \frac{dx}{1-\left(\frac{1-ta{n}^2\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}\right)-\frac{2tan\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}}$

$=\int \frac{se{c}^2\frac{x}{2}dx}{2ta{n}^2\frac{x}{2}-2tan\frac{x}{2}}=\int \frac{\frac{1}{2}se{c}^2\frac{x}{2}dx}{ta{n}^2\frac{x}{2}-tan\frac{x}{2}}$

Put $tan\frac{x}{2}=t\Rightarrow \frac{1}{2}se{c}^2\frac{x}{2}dx=dt$

$\therefore I=\int \frac{dt}{t^2-t}=\left[\frac{1}{t-1}-\frac{1}{t}\right]\int +C$

$=log\left(t-1\right)-logt+C=log\left|\frac{t-1}{t}+C\right|$

$=log\left|1-cot\frac{x}{2}\right|+C$