Question

\[\int_0^\pi {\dfrac{1}{{\left[ {1{\text{ }} + {\text{ }}sin{\text{ }}x} \right]}}{\text{ }}dx} \]
A. 1
B. 2
C. -1
D. -2

Answer 1

Hint: In this question we have to find the integral value of \[\int_0^\pi {\dfrac{1}{{\left[ {1{\text{ }} + {\text{ }}sin{\text{ }}x} \right]}}{\text{ }}dx} \]. To solve these types of questions we must know the basic integral formula used for sin, cos, and tan, and often we can derive many formulas from a single one which makes the question to solve as we have done in this question.

Formula Used:
 \[1 + \sin x = \left[ {\cos \dfrac{x}{2} + \sin\dfrac{x}{2}} \right]\]
\[\left[ {\cos \dfrac{x}{2} + sin\dfrac{x}{2}} \right] = \left[ {1 + \tan \dfrac{x}{2}} \right]\]

Complete step by step Solution:
Given: \[\int_0^\pi {\dfrac{1}{{\left[ {1{\text{ }} + {\text{ }}sin{\text{ }}x} \right]}}{\text{ }}dx} \]
LET \[I = \int_0^\pi {\dfrac{1}{{\left[ {1{\text{ }} + {\text{ }}sin{\text{ }}x} \right]}}{\text{ }}dx} \]
With the help of integral property as we have taken $\pi $ to $\dfrac{\pi }{2}$ so the same goes for x
By using the formula \[1 + \sin x = \left[ {\cos \dfrac{x}{2} + sin\dfrac{x}{2}} \right]\]in equation (1)
\[ = \int_0^{\dfrac{\pi }{2}} {\dfrac{1}{{{{\left[ {\cos \dfrac{x}{2} + sin\dfrac{x}{2}} \right]}^2}}}{\text{ }}dx} \]
Now, by using \[\left[ {\cos \dfrac{x}{2} + sin\dfrac{x}{2}} \right] = \left[ {1 + \tan \dfrac{x}{2}} \right]\] and using it in equation (2)
\[I = \int_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sec }^2}\dfrac{x}{2}}}{{{{\left[ {1 + \tan \dfrac{x}{2}} \right]}^2}}}{\text{ }}dx} \]
\[{\text{ }}\]After simplifying the above equation we get
$\tan \dfrac{x}{2} = t$
$dt = \dfrac{1}{2}{\sec ^2}\dfrac{x}{2}dx$
Therefore, $\int_0^\infty {\dfrac{{2dt}}{{{{\left( {1 + t} \right)}^2}}}} $
$ = \left[ {\dfrac{{ - 2}}{{1 + t}}} \right]_0^\infty $
= 2

Hence, the correct option is B.

Note: In this type of question where we integral value needs to be found out main aim is to make them or divide them into those easy formulas which are very versatile in nature and can be easily solved. Like in this question where we took cos and sin into a tan form which was easier to solve.