Question

Evaluate: ${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{dx}{(1+\sin x){\sec }^{2}x}$

• A. $\pi$
• B. $2\pi$
• C. $0$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is A $\pi$

Let, $I={\int }_{-\pi /2}^{\pi /2}\frac{dx}{(1+\sin x)se{c}^{2}x}=I_1\left(say\right)$

Using the property:-

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$⟹I={\int }_{-\pi /2}^{\pi /2}\frac{dx}{(1+\sin (\pi /2-\pi /2-x))se{c}^2(\pi /2-\pi /2-x)}=I_2\left(say\right)$

$={\int }_{-\pi /2}^{\pi /2}\frac{dx}{(1-\sin x)se{c}^{2}x}$

Now, $2I=I_1+I_2$

$⟹2I={\int }_{-\pi /2}^{\pi /2}\frac{2dx}{(1-{\sin }^{2}x){\sec }^{2}x}$

$={\int }_{-\pi /2}^{\pi /2}\frac{2dx}{{\cos }^{2}x.{\sec }^{2}x}={\int }_{-\pi /2}^{\pi /2}2dx$

$⟹2I=2{\left[x\right]}_{-\pi /2}^{\pi /2}=2\pi$

$⟹I=\pi$

Hence, answer is option-(A).