Question

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

Answer 1

Hint: 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.