Question

$\int \frac{x-\sin x}{1-\cos x}dx=$

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

Answer 1

The correct option is D $-x\cot \frac{x}{2}+c$

$\int \left(\frac{x-sinx}{1-cosx}\right)dx=\int \frac{x-sinx}{2si{n}^{2}x/2}dx$

$=\int \left(\frac{x}{2}\right)cose{c}^{2}x/2^{dx}-\int cotx/2dx$

$=x/2\int cose{c}^{2}x/2d(x/2-\int (\int cose{c}^{2}x/2dx/2\left)\right)dx$

$-\int cotx/2dx$

$=-xcotx/2+c+\int cotx/2dx-\int cotx/2dx$

$=-xcotx/2+c$

$\int \left(\frac{x-sinx}{1-cosx}\right)dx=-xcotx/2+c$