Question

Solve $\underset{\frac{\pi }{2}}{\overset{\pi }{\int }}\frac{1-\sin x}{1-\cos x}dx$

• A. $1-2\log 2$
• B. $1-\log 2$
• C. $1+\log 2$
• D. None of these

Answer 1

The correct option is B $1-\log 2$

Given

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

We know that

$\sin x=2\cos \frac{x}{2}\sin \frac{x}{2}$ -- (1)

&

$\cos x=1-2{\sin }^2\frac{x}{2}$

So

$1-\cos x=1-(1-2{\sin }^2\frac{x}{2})=2{\sin }^2\frac{x}{2}$ -- (2)

Put values of (1) & (2) in the expression

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

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

$=\frac{1}{2}{\int }_{\frac{\pi }{2}}^{\pi }{\left({\csc }^2\frac{x}{2}\right)}^{dx}-{\int }_{\frac{\pi }{2}}^{\pi }{\left(\cot \frac{x}{2}\right)}^{dx}$

Now we know

$\int {\csc }^2\frac{x}{2}dx=-\cot \frac{x}{2}.2+C$

$\int \cot \frac{x}{2}dx=log\left|\sin \frac{x}{2}\right|.2+C$

Using above formulas, we get

$=\frac{1}{2}{\int }_{\frac{\pi }{2}}^{\pi }\left({\csc }^2\frac{x}{2}\right)dx-{\int }_{\frac{\pi }{2}}^{\pi }\left(\cot \frac{x}{2}\right)dx$

$=\frac{1}{2}(-\cot \frac{x}{2}).2\stackrel{\pi }{\underset{\frac{\pi }{2}}{|}}-(\log \left|\sin \frac{x}{2}\right|\times 2)\stackrel{\pi }{\underset{\frac{\pi }{2}}{|}}$

$=-\cot \frac{x}{2}\stackrel{\pi }{\underset{\frac{\pi }{2}}{|}}-2\log \left|\sin \frac{x}{2}\right|\stackrel{\pi }{\underset{\frac{\pi }{2}}{|}}$

$=-\left(\cot \frac{\pi }{2}-\cot \frac{\pi }{4}\right)-2\left(\log \left(\sin \frac{\pi }{2}\right)-\left(\sin \frac{\pi }{4}\right)\right)$

$=-(0-1)-2(\log 1-\log \frac{1}{\sqrt{2}})$

$=1-2(0-\log 2^{-\frac{1}{2}})$

$=1-2(-.(-\frac{1}{2})\log \left(2\right))$

$=1-2.\frac{1}{2}\log 2$

$=1-\log 2$

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