Question

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

• A. ${\int }_0^{log(1+\sqrt{2})}2(e^u+e^{-u}{)}^{16}du$
• B. ${\int }_0^{log(1+\sqrt{2})}2(e^u+e^{-u}{)}^{17}du$
• C. ${\int }_0^{log(1+\sqrt{2})}2(e^u-e^{-u}{)}^{17}du$
• D. ${\int }_0^{log(1+\sqrt{2})}2(e^u-e^{-u}{)}^{16}du$

Answer 1

The correct option is A ${\int }_0^{log(1+\sqrt{2})}2(e^u+e^{-u}{)}^{16}du$

${\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}(2cosecx{)}^{17}dx$

Let
$e^u+e^{-u}=2cosecx$

$x=\frac{\pi }{4}\Rightarrow u=ln(1+\sqrt{2})$

$x=\frac{\pi }{2}\Rightarrow u=0$
$\Rightarrow cosecx+cotx=e^u$ and $cosecx-cotx=e^{-u}$

$\Rightarrow x=\frac{e^u-e^{-u}}{2}(e^u-e^{-u})dx=-2cosecxcotxdx$

$\Rightarrow -\int (e^u+e^{-u}{)}^{17}\frac{(e^u-e^{-u})}{2cosecxcotx}du$

$=-2{\int }_{ln(1+\sqrt{2})}^0(e^u+e^{-u}{)}^{16}du$

$={\int }_0^{ln(1+\sqrt{2})}2(e^u+e^{-u}{)}^{16}du$