Question

Evaluate: ${\int }_{\frac{\pi }{2}}^{\pi }\frac{1-sinx}{1-cosx}dx$

Answer 1

${\int }_{\frac{\pi }{2}}^{\pi }\frac{1-sinx}{1-cosx}dx=\int \frac{1}{1-\cos x}dx-\int \frac{\sin x}{1-\cos x}dx=\int \frac{1}{2{\sin }^2\frac{x}{2}}dx-\int \frac{\sin x}{1-\cos x}dx=\frac{1}{2}\int cose{c}^2\frac{x}{2}-\int \frac{\sin x}{1-\cos x}dx=\frac{1}{2}\ast 2\ast {\left[\cot \frac{x}{2}\right]}_{\frac{\pi }{2}}^{\pi }-\int \frac{\sin x}{1-\cos x}dx=[\cot \frac{\pi }{2}-\cot \frac{\pi }{4}]-\int \frac{\sin x}{1-\cos x}dx=0-1-\int \frac{\sin x}{1-\cos x}dx$

Now, for

${\int }_{\frac{\pi }{2}}^{\pi }\frac{\sin x}{1-\cos x}dx$

Let $1-\cos x=t\Rightarrow \sin xdx=dt$

$\Rightarrow \int \frac{dt}{t}=\log t+c={\left[\log \right(1-\cos x\left)\right]}_{\frac{\pi }{2}}^{\pi }=\log 2$

hence, ${\int }_{\frac{\pi }{2}}^{\pi }\frac{1-sinx}{1-cosx}dx=0-1-\int \frac{\sin x}{1-\cos x}dx=-1-\log 2$