JEE Main 2022 — Mathematics Calculus

$\frac{dy}{dx}+\frac{xy}{{x}^{2}-1}=\frac{{x}^{4}+2x}{\sqrt{1-{x}^{2}}}$dxdy​+x2−1xy​=1−x2​x4+2x​ is a linear differential equation.

Here $\displaystyle I.F.={e}^{\int \frac{x}{{x}^{2}-1}dx}={e}^{\frac{1}{2}\int \frac{2x}{{x}^{2}-1}dx}={e}^{\frac{1}{2}\mathrm{ln}({x}^{2}-1)}$I.F.=e∫x2−1x​dx=e21​∫x2−12x​dx=e21​ln(x2−1)

i.e. $I.F.=\sqrt{1-{x}^{2}}$I.F.=1−x2​$\quad \because x \in(-1,1)$∵x∈(−1,1)

So, solution of the differential equation is

$$y\cdot \sqrt{1-{x}^{2}}=\int \frac{{x}^{4}+2x}{\sqrt{1-{x}^{2}}}\cdot \sqrt{1-{x}^{2}}dx$$y⋅1−x2​=∫1−x2​x4+2x​⋅1−x2​dx

$$y\sqrt{1-{x}^{2}}=\int ({x}^{4}+2x)dx$$y1−x2​=∫(x4+2x)dx

$$y\sqrt{1-{x}^{2}}=\frac{{x}^{5}}{5}+{x}^{2}+C$$y1−x2​=5x5​+x2+C

Since the curve passes through origin

so $x=0,y=0$x=0,y=0 $\Rightarrow C=0$⇒C=0

i.e. $y=\frac{{x}^{5}}{5\sqrt{1-{x}^{2}}}+\frac{{x}^{2}}{\sqrt{1-{x}^{2}}}$y=51−x2​x5​+1−x2​x2​

Now,

$${\int }_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}f(x)dx={\int }_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}\frac{{x}^{5}}{5\sqrt{1-{x}^{2}}}dx+{\int }_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}\frac{{x}^{2}}{\sqrt{1-{x}^{2}}}dx$$∫2−3​​23​​​f(x)dx=∫2−3​​23​​​51−x2​x5​dx+∫2−3​​23​​​1−x2​x2​dx

$\displaystyle {\int }_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}f(x)dx=0+2{\int }_{0}^{\frac{\sqrt{3}}{2}}\frac{{x}^{2}}{\sqrt{1-{x}^{2}}}dx$∫2−3​​23​​​f(x)dx=0+2∫023​​​1−x2​x2​dx (as $\frac{{x}^{5}}{5\sqrt{1-{x}^{2}}}$51−x2​x5​ is an odd function and$\frac{{x}^{2}}{\sqrt{1-{x}^{2}}}$1−x2​x2​ is an even function)

Let $\displaystyle I=\int \frac{{x}^{2}}{\sqrt{1-{x}^{2}}}dx$I=∫1−x2​x2​dx

Assume $1-{x}^{2}={t}^{2}\Rightarrow 1-{t}^{2}={x}^{2}$1−x2=t2⇒1−t2=x2

and $-2tdt=2xdx\Rightarrow dx=-\frac{tdt}{\sqrt{1-{t}^{2}}}$−2tdt=2xdx⇒dx=−1−t2​tdt​

i.e. $\displaystyle I=\int \frac{1-{t}^{2}}{t}(-\frac{t}{\sqrt{1-{t}^{2}}})dt$I=∫t1−t2​(−1−t2​t​)dt

$$=-\int \sqrt{1-{t}^{2}}dt$$=−∫1−t2​dt

$$=-[\frac{t}{2}\sqrt{1-{t}^{2}}+\frac{1}{2}{\mathrm{sin}}^{-1}t]+C$$=−[2t​1−t2​+21​sin−1t]+C

$$=-[\frac{x}{2}\sqrt{1-{x}^{2}}+\frac{1}{2}{\mathrm{sin}}^{-1}\sqrt{1-{x}^{2}}]+C$$=−[2x​1−x2​+21​sin−11−x2​]+C

$$\Rightarrow {\int }_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}f(x)dx$$⇒∫2−3​​23​​​f(x)dx

$$=2{[-\frac{x}{2}\sqrt{1-{x}^{2}}-\frac{1}{2}{\mathrm{sin}}^{-1}\sqrt{1-{x}^{2}}]}_{0}^{\frac{\sqrt{3}}{2}}$$=2[−2x​1−x2​−21​sin−11−x2​]023​​​

$$=2(-\frac{\sqrt{3}}{8}-\frac{1}{2}{\mathrm{sin}}^{-1}\frac{1}{2}+\frac{1}{2}{\mathrm{sin}}^{-1}1)$$=2(−83​​−21​sin−121​+21​sin−11)

$$=-\frac{\sqrt{3}}{4}-\frac{\pi }{6}+\frac{\pi }{2}=\frac{\pi }{3}-\frac{\sqrt{3}}{4}$$=−43​​−6π​+2π​=3π​−43​​