Question

Evaluate : ${\int }_0^{\pi }\frac{1}{\left[1+\sin x\right]}dx$

• A. $1$
• B. $2$
• C. $-1$
• D. $-2$

Answer 1

The correct option is B

$2$

Consider the given equation as,

$I={\int }_0^{\mathrm{\pi }}\frac{1}{\left[1+\sin x\right]}dx$

Rewrite the above Equation by using the rationalization method.

$$\begin{gathered} I={\int }_0^{\pi }\frac{1}{\left[1+\sin x\right]}dx\times \frac{1-\sin x}{1-\sin x}dx \\ I={\int }_0^{\mathrm{\pi }}\frac{1-\sin x}{1-{\sin }^{2}x}dx \\ I={\int }_0^{\pi }\frac{1-\sin x}{{\cos }^{2}x}dx \\ I={\int }_0^{\mathrm{\pi }}\left[\frac{1}{{\cos }^{2}x}-\frac{\sin x}{\cos x}\times \frac{1}{\cos x}\right]dx \\ I={\int }_0^{\mathrm{\pi }}\left(se{c}^{2}x-secx.\tan x\right)dx \\ I={\left[\tan x-secx\right]}_0^{\mathrm{\pi }} \\ I=\left(\tan \mathrm{\pi }-\sec \mathrm{\pi }\right)-\left(\tan 0-sec0\right) \\ I=\left[\left(0\right)-\left(-1\right)\right]-\left[\left(0\right)-\left(1\right)\right] \\ I=1-\left(-1\right) \\ I=2 \end{gathered}$$

Hence, the correct answer is Option B.