Question

Evaluate the integral $\underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{x\sin x}{1+\sin x}}dx$.

Answer 1

Hint: Solve the integral by replacing x by $(\pi -x)$as per $\underset{0}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx$. Then simplify it using trigonometric identities. Finally, after integration substitute $(\pi ,0)$in the place of x.

Complete step-by-step solution -
Given the integral, $\underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{x\sin x}{1+\sin x}}dx$.
Let’s put, $I=\underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{x\sin x}{1+\sin x}}dx$.
We know that, $\underset{0}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx$.
Thus, x becomes $(\pi -x)$.
$\therefore I=\underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{(\pi -x)\sin x}{1+\sin (\pi -x)}}dx$
We know, $$\sin (180-\theta )=\sin \theta$$
                   $$\sin (\pi -x)=\sin x$$
$$\begin{array}{rl}& I=\underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{(\pi -x)\sin x}{1+\sin x}}=\underset{0}{\overset{\pi }{\int }}\left({\displaystyle \frac{\pi \sin x-x\sin x}{1+\sin x}}\right)dx\\ & I=\underset{0}{\overset{\pi }{\int }}({\displaystyle \frac{\pi \sin x}{1+\sin x}}-{\displaystyle \frac{x\sin x}{1+\sin x}})dx\end{array}$$
$$I=\underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{\pi \sin x}{1+\sin x}}dx-\underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{x\sin x}{1+\sin x}}dx$$
$$I=\underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{\pi \sin x}{1+\sin x}}dx-I$$
$$\begin{array}{rl}& \Rightarrow I+I=\underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{\pi \sin x}{1+\sin x}}dx\\ & 2I=\underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{\pi \sin x}{1+\sin x}}dx\\ & \therefore I={\displaystyle \frac{\pi }{2}}\underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{\sin x}{1+\sin x}}dx\end{array}$$
Multiply numerator and denominator with $$(1-\sin x)$$.
$$\begin{array}{rl}& I={\displaystyle \frac{\pi }{2}}\underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{\sin x(1-\sin x)}{(1+\sin x)(1-\sin x)}}dx\\ & ={\displaystyle \frac{\pi }{2}}\underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{\sin x(1-\sin x)}{1-{\sin }^{2}x}}dx\\ & ={\displaystyle \frac{\pi }{2}}\underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{\sin x-{\sin }^{2}x}{1-{\sin }^{2}x}}dx\\ & ={\displaystyle \frac{\pi }{2}}\underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{\sin x}{{\cos }^{2}x}}dx-{\displaystyle \frac{\pi }{2}}\underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{{\sin }^{2}x}{{\cos }^{2}x}}dx\\ & ={\displaystyle \frac{\pi }{2}}\underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{\tan x}{\cos x}}dx-{\displaystyle \frac{\pi }{2}}\underset{0}{\overset{\pi }{\int }}{\tan }^{2}xdx\\ & ={\displaystyle \frac{\pi }{2}}\underset{0}{\overset{\pi }{\int }}\tan x.\sec x.dx-{\displaystyle \frac{\pi }{2}}\underset{0}{\overset{\pi }{\int }}{\tan }^{2}x.dx\end{array}$$
We know, $$(a-b)(a+b)=a^2-b^2$$.
$$\begin{array}{rl}& {\sin }^{2}x+{\cos }^{2}x=1\\ & \therefore {\cos }^{2}x=1-{\sin }^{2}x\\ & \tan x={\displaystyle \frac{\sin x}{\cos x}}\end{array}$$
Which are basic, trigonometric formulae.
$$\because {\displaystyle \frac{1}{\cos x}}=\sec x$$
We know $$\int \tan x.\sec x=\sec x$$and $$\int {\sec }^{2}x=\tan x$$.
Similarly, $${\tan }^{2}x={\sec }^{2}x-1$$.
$$\begin{array}{rl}& ={\displaystyle \frac{\pi }{2}}\underset{0}{\overset{\pi }{\int }}\tan x.\sec x.dx-{\displaystyle \frac{\pi }{2}}\underset{0}{\overset{\pi }{\int }}({\sec }^2-1)dx\\ & ={\displaystyle \frac{\pi }{2}}{[\sec x]}_0^{\pi }-{\displaystyle \frac{\pi }{2}}[\underset{0}{\overset{\pi }{\int }}{\sec }^{2}xdx-\underset{0}{\overset{\pi }{\int }}1.dx]\\ & ={\displaystyle \frac{\pi }{2}}{[\sec x]}_0^{\pi }-{\displaystyle \frac{\pi }{2}}[{[\tan x]}_0^{\pi }-{\left[x\right]}_0^{\pi }]\\ & ={\displaystyle \frac{\pi }{2}}[{[\sec x]}_0^{\pi }-{[\tan x]}_0^{\pi }+{\left[x\right]}_0^{\pi }]\\ & ={\displaystyle \frac{\pi }{2}}[(\sec \pi -\sec 0)-(\tan \pi -\tan 0)+(\pi -0)]\end{array}$$
$$\sec \pi =-1$$ $$\tan \pi =0$$
$$\sec 0=+1$$ $$\tan 0=0$$
$$\begin{array}{rl}& ={\displaystyle \frac{\pi }{2}}[[-1-1]+\pi ]\\ & ={\displaystyle \frac{\pi }{2}}[-2+\pi ]\\ & ={\displaystyle \frac{\pi (\pi -2)}{2}}\\ & \therefore I={\displaystyle \frac{\pi (\pi -2)}{2}}\end{array}$$
Hence, by evaluating the integral, we get $${\displaystyle \frac{\pi (\pi -2)}{2}}$$.
Note:- Be careful while simplifying the integral. Open brackets, don’t mix up the sign. Remember the basic identities and trigonometric formulae. You should learn the integral values of $$\tan x.\sec x,{\sec }^{2}x$$ etc, which we have used in solving the integral. Finally substitute $$(\pi ,0)$$and simplify the expression.