Question

The following integral $\underset{\pi /4}{\overset{\pi /2}{\int }}(2cosecx{)}^{17}dx$ is equal to

• A. $\underset{0}{\overset{\log (1+\sqrt{2})}{\int }}2(e^u+e^{-u}{)}^{16}du$
• B. $\underset{0}{\overset{\log (1+\sqrt{2})}{\int }}(e^u+e^{-u}{)}^{17}du$
• C. $\underset{0}{\overset{\log (1+\sqrt{2})}{\int }}(e^u-e^{-u}{)}^{17}du$
• D. $\underset{0}{\overset{\log (1-\sqrt{2})}{\int }}2(e^u-e^{-u}{)}^{16}du$

Answer 1

Answer: (A)

$\underset{\frac{\pi }{4}}{\overset{\frac{\pi }{2}}{\int }}(2\mathrm{cosec}x{)}^{17}dx$
Let $e^u+e^{-u}=2\mathrm{cosec}x,x=\frac{\pi }{4}$
$\Rightarrow u=\ln (1+\sqrt{2}),x=\frac{\pi }{2}$
$\Rightarrow u=0$
$\Rightarrow \mathrm{cosec}x+\cot x=e^u$
and $\mathrm{cosec}x-\cot x=e^{-u}$
$\Rightarrow \cot x=\frac{e^u-e^{-u}}{2}$
$\left(e^u-e^{-u}\right)dx=-2\mathrm{cosec}x\cot xdx$
$\Rightarrow -\int {\left(e^u+e^{-u}\right)}^{17}\frac{\left(e^u-e^{-u}\right)}{2\mathrm{cosec}x\cot x}du$
$=-2\underset{\ln (1+\sqrt{2})}{\overset{0}{\int }}{\left(e^u+e^{-u}\right)}^{16}du$
$=\underset{0}{\overset{\ln (1+\sqrt{2})}{\int }}2{\left(e^u+e^{-u}\right)}^{16}du$