Question

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

• A. $-\log \left(2\right)-1$
• B. $-\log \left(2\right)+1$
• C. $\frac{1}{\log \left(2\right)}+1$
• D. $\frac{1}{\log \left(2\right)}-1$

Answer 1

The correct option is B $-\log \left(2\right)+1$

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

$={\int }_{\frac{\pi }{2}}^{\pi }\frac{1-2\sin \frac{x}{2}.\cos \frac{x}{2}}{2{\sin }^2\frac{x}{2}}.dx[\because \sin 2x=2\sin x.\cos x,\cos 2x=1-2{\sin }^{2}x]$

$={\int }_{\frac{\pi }{2}}^{\pi }(\frac{1}{2}cose{c}^2\frac{x}{2}-\cot \frac{x}{2}).dx$

$=2.\frac{1}{2}{[-\cot z]}_{\frac{\pi }{4}}^{\frac{\pi }{4}}-2[\log |\sin x|{]}_{\frac{\pi }{4}}^{\frac{\pi }{2}}=[-(0-1)]-2[\log 1-\log \left(\frac{1}{\sqrt{2}}\right)]$

$=1-2[0+\log \sqrt{2}][\because \log (1/x)=-\log x,\log (x^n)=n\log x]$

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