Question

${\int }_{\pi /4}^{\pi /2}\cos \theta {\csc }^2\theta d\theta =$

• A. $\sqrt{2}-1$
• B. $1-\sqrt{2}$
• C. $\sqrt{2}+1$
• D. $Noneofthese$

Answer 1

The correct option is A $\sqrt{2}-1$

${\int }_{\pi /4}^{\pi /2}\frac{\cos \theta }{{\sin }^2\theta }$

$\sin \theta =t$

$\cos \theta d\theta =dt$

${\int }_{\pi /4}^{\pi /2}{t}^{-2}dt={\left(\frac{t-1}{-1}\right)}_{\pi /4}^{\pi /2}={-\left(\frac{1}{t}\right)}_{\pi /4}^{\pi /2}={(+\frac{1}{\sin \theta })}_{\pi /4}^{\pi /2}$

$=-(1-\sqrt{2})=\sqrt{2}-1$