Question

The 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

Answer: (A)

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

We know;

$\int cosecxdx=\log |cosecx+cotx|$

$I={\int }_{\pi /4}^{\pi /2}(2cosecx{)}^{17}dx=$ $I={\int }_{\pi /4}^{\pi /2}(2cosecx{)}^{16}cosecxdx$

let $u=\log |cosecx+cotx|$

at $x=\pi /4,u=0$ and at $x=\pi /2,u=\log (1+\sqrt{2})$

$⟹du=cosecxdx$

$e^u=cosecx+cotx$

$e^{-u}=cosecx-cotx$

because $e^u{e}^{-u}=cose{c}^{2}x-co{t}^{2}x=1$

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