Question

${\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{\sqrt{1+cosx}}{(1-cosx{)}^{\frac{5}{2}}}dx=$ [AI CBSE 1980]

• A. $\frac{5}{2}$
• B. $\frac{3}{2}$
• C. $\frac{1}{2}$
• D. $\frac{2}{5}$

Answer 1

The correct option is B

$\frac{3}{2}$

$I={\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{\sqrt{1+cosx}}{(1-cosx{)}^{\frac{5}{2}}}\times \frac{\sqrt{1-cosx}}{1-cosx}dx$
$={\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{sinx}{(1-cosx{)}^3}dx$
Now, put 1 - cos~x=t
Also, when $x=\frac{\pi }{3},t=\frac{1}{2}andx=\frac{\pi }{2},t=1$
Therefore, $I={\int }_{\frac{1}{2}}^1\frac{dt}{t^3}={\left|\frac{t^{-2}}{-2}\right|}_{\frac{1}{2}}^1=\frac{3}{2}$