Question

Integration of $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x}{\sin x+cosx}dx}$?

Answer 1

Hint: Here we have to integrate the given definite integral value. Firstly we will write the properties of a definite integral $\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$ and compare it with the integral given to us and form a new value. Then we will add the value given and the new value such that most of the terms are cancelled out. Finally we will integrate the value obtained and get our desired answer.

Complete step by step solution:
We have to integrate the value given below:
$\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x}{\sin x+\cos x}dx}$
Let the above value be equal to some Variable $I$ as follows:
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x}{\sin x+cosx}dx}$….$\left( 1 \right)$
Comparing the above value from the properties of definite integral below:
$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$….$\left( 2 \right)$
We get,
$f\left( x \right)=\dfrac{\sin x}{\sin x+\cos x}$
$a=\dfrac{\pi }{2}$
So,
$f\left( a-x \right)=f\left( \dfrac{\pi }{2}-x \right)$
$\Rightarrow f\left( \dfrac{\pi }{2}-x \right)=\dfrac{\sin \left( \dfrac{\pi }{2}-x \right)}{\sin \left( \dfrac{\pi }{2}-x \right)+\cos \left( \dfrac{\pi }{2}-x \right)}$
Now as we know $\sin \left( {{90}^{\circ }}-x \right)=\cos x$ and $\cos \left( {{90}^{\circ }}-x \right)=\sin x$ as all trigonometric function are positive in first quadrant replace it above,
$\Rightarrow f\left( \dfrac{\pi }{2}-x \right)=\dfrac{\cos x}{\cos x+\sin x}$
Using the equation (2) we can replace the above value in equation (1) follows:
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x}{\cos x+\sin x}dx}$….$\left( 3 \right)$
Add equation (1) and (3),
$I+I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x}{\sin x+cosx}dx}+\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x}{\cos x+\sin x}dx}$
$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x+\cos x}{\sin x+cosx}dx}$
Simplifying further we get,
$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{1dx}$
We can write $1={{x}^{0}}$ and using the formula $\int\limits_{a}^{b}{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ we get,
$\Rightarrow 2I=\left[ x \right]_{0}^{\dfrac{\pi }{2}}$
$\Rightarrow I=\dfrac{1}{2}\left[ \dfrac{\pi }{2}-0 \right]$
So we get,
$\Rightarrow I=\dfrac{\pi}{4}$
Hence Integration of $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x}{\sin x+cosx}dx}$ is $\dfrac{\pi }{4}$ .

Note:
Integration is an inverse operation of differentiation. There are two types of integral, one whose limits are not given known as the indefinite integral and others whose limits are given known as definite integral. As we have been given a Definite integral we have not added constant in the final answer. We can do this question without using the property of definite integration by rationalizing the denominator and simplifying it further but that will be a lengthy way. So use the properties until it’s not stated by which method we have to solve the problem.