Answer:
Option B
Explanation:
Plan
(i) solution of the differential equation
$\frac{dy}{dx}+Py=Q$ is
$Y.(IF)=\int_{}^{} Q.(IF) dx+c$
where,IF=$e^{\int P dx}$
(ii)$\int_{-a}^{a }f(x)dx=2\int_{0}^{a} f(x)dx,$
if f(-x)= f(x)
Given differential equation
$\frac{dy}{dx}+\frac{x}{x^{2}-1}y=\frac{x^{4}+2x}{\sqrt{1-x^{2}}}$
This is a linear differential solution
IF= $e^{\int \frac{x}{x^{2}-1}dx}=e^{\frac{1}{2}ln|x^{2}-1|}$
= $\sqrt{1-x^{2}}$
$\Rightarrow$ Solution is
$y\sqrt{1-x^{2}}=\int_{}^{} \frac{x(x^{3}+2)}{\sqrt{1-x^{2}}}.\sqrt{1-x^{2}}dx$
or $y\sqrt{1-x^{2}}=\int_{}^{} (x^{4}+2x) dx$
$=\frac{x^{5}}{5}+x^{2}=c$
f(0)=0 $\Rightarrow$ c=0
$\Rightarrow$ $f(x)\sqrt{1-x^{2}}=\frac{x^{5}}{5}+x^{2}$
Now, $\int_{-\sqrt{3}/2}^{\sqrt{3}/2}f(x) dx= \int_{-\sqrt{3}/2}^{\sqrt{3}/2}\frac{x^{2}}{\sqrt{1-x^{2}}}dx$
[using property]
$=2\int_{0}^{\sqrt{3}/2}\frac{x^{2}}{\sqrt{1-x^{2}}}dx$
$=2\int_{0}^{{\pi}/3}\frac{\sin^{2}\theta}{\cos\theta}\cos\theta d\theta$
[ taking x= sin $\theta$]
$=2\int_{0}^{{\pi}/3}\sin^{2}\theta d\theta$
$=\int_{0}^{{\pi}/3}(1-cos2\theta) d\theta$
= $\left(\theta-\frac{\sin 2\theta}{2}\right)_{0}^{\pi/3}$
= $\frac{\pi}{3}-\frac{\sin 2\pi/3}{2}$
= $\frac{\pi}{3}-\frac{\sqrt{3}}{4}$
