Question

Evaluate ${\int }_0^{\frac{\pi }{2}}\frac{\cos x}{1+{\sin }^{2}x}dx$:

• (A) ${90}^{\circ }$
• (B) ${45}^0$
• (C) ${\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)$
• (D) $-{\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Answer 1

The Correct Option is B

Explanation:
Substitution method: If the function cannot be integrated directly substitution method is used. To integration by substitution is used in the following steps:A new variable is to be chosen, say “t”The value of dt is to is to be determined.Substitution is done and integral function is then integrated.Finally, initial variable t, to be returned.Let $\sin x=t\Rightarrow \cos xdx=dt$$\Rightarrow 1=\int \frac{\cos x}{1+{\sin }^{2}x}dx$$\Rightarrow 1=\int \frac{1}{1+t^2}dt$$\Rightarrow 1={\tan }^{-1}t+c$$\Rightarrow 1={\tan }^{-1}(\sin x)+c$Putting the limits,$\Rightarrow 1=[{\tan }^{-1}(\sin {90}^{\circ })+c]-[{\tan }^{-1}(\sin 0^{\circ })+c]$$\Rightarrow 1={\tan }^{-1}(1)$$\Rightarrow I={45}^{\circ }$ or $\frac{\pi }{4}$Hence, the correct option is (B).