Question

Evaluate the given integration $\int\limits_{0}^{2\pi }{\dfrac{1}{1+{{e}^{\sin x}}}}$

Answer 1

To find the value of $\int\limits_{0}^{2\pi }{\dfrac{1}{1+{{e}^{\sin x}}}}$ :this is a definite integral of limits from 0 to $2\pi$ so we will use the property of definite integrals. $\int\limits_{0}^{2a}{f(x)}=\int\limits_{0}^{a}{f(x)}+\int\limits_{0}^{a}{f(2\pi -x)}$ and then after solving a little and using one trigonometric property $\sin (2\pi -x)=-sinx$ and then our whole expression converts to 1 and then integrate 1 from 0 to $\pi$ , we get the answer as $\pi$

Complete step-by-step solution:
We have to find value of $\int\limits_{0}^{2\pi }{\dfrac{1}{1+{{e}^{\sin x}}}}$ , so before going onto this let us remind one property of definite integral that is $\int\limits_{0}^{2a}{f(x)}=\int\limits_{0}^{a}{f(x)}+\int\limits_{0}^{a}{f(2\pi -x)}$ , now let’s compare this property to our question
We can compare a with $\pi$ ,so we can write our expression as $\int\limits_{0}^{2\pi }{\dfrac{1}{1+{{e}^{\sin x}}}}dx=\int\limits_{0}^{\pi }{\dfrac{1}{1+{{e}^{\sin x}}}dx}+\int\limits_{0}^{\pi }{\dfrac{1}{1+{{e}^{\sin (2\pi -x)}}}dx}$
So now we have to solve RHS value only now recalling one property of trigonometry that is
$\sin (2\pi -x)=-sinx$, it means we can write ${{e}^{\sin (2\pi -x)}}={{e}^{-\sin (x)}}$ so on applying this property on our LHS we can write it as
$\int\limits_{0}^{2\pi }{\dfrac{1}{1+{{e}^{\sin x}}}}dx=\int\limits_{0}^{\pi }{\dfrac{1}{1+{{e}^{\sin x}}}dx}+\int\limits_{0}^{\pi }{\dfrac{1}{1+{{e}^{-\sin x}}}dx}$, now we can also write this expression as
Using this property $\int\limits_{a}^{b}{f(x)\pm g(x)}=\int\limits_{a}^{b}{f(x)\pm \int\limits_{a}^{b}{g(x)}}$
$\int\limits_{0}^{2\pi }{\dfrac{1}{1+{{e}^{\sin x}}}}dx=\int\limits_{0}^{\pi }{(\dfrac{1}{1+{{e}^{\sin x}}}}+\dfrac{1}{1+{{e}^{-\sin x}}})dx........(1)$ because the limits are same
${{e}^{-x}}=\dfrac{1}{{{e}^{x}}}$ similarly, ${{e}^{-\sin x}}=\dfrac{1}{{{e}^{\sin x}}}$
Now on solving we can write expression $\dfrac{1}{1+{{e}^{-\sin x}}}=\dfrac{{{e}^{\sin x}}}{1+{{e}^{\sin x}}}.......(2)$
Now on putting equation (2) back into the equation (1)
We get expression $\int\limits_{0}^{2\pi }{\dfrac{1}{1+{{e}^{\sin x}}}dx}=\int\limits_{0}^{\pi }{(\dfrac{1}{1+{{e}^{\sin x}}}}+\dfrac{{{e}^{\sin x}}}{1+{{e}^{\sin x}}})dx$
Which can be further written as $\int\limits_{0}^{2\pi }{\dfrac{1}{1+{{e}^{\sin x}}}}dx=\int\limits_{0}^{\pi }{\dfrac{1+{{e}^{\sin x}}}{1+{{e}^{\sin x}}}}dx=\int\limits_{0}^{\pi }{1}dx$
we get $[x]_{0}^{\pi }$
Putting the limits, we get answer $\pi -0$ , hence answer is $\pi$

Note: Integral like this $\int\limits_{0}^{2\pi }{\dfrac{1}{1+{{e}^{\sin x}}}}dx$ (we have $\sin x$ in place of x ) can never be solved straight , so always try to use property of definite integrals at first for example $\int\limits_{a}^{b}{f(x)\pm g(x)}=\int\limits_{a}^{b}{f(x)\pm \int\limits_{a}^{b}{g(x)}}$ and $\int\limits_{a}^{c}{f(x)=\int\limits_{a}^{b}{f(x)+\int\limits_{b}^{c}{f(x)}}}$
Then it automatically resolves into some simple problem and can be solved easily, the same as we did in this question.