© All Rights reserved @ LearnWithDash
Step 1: Identify the Type of Differential Equation
The given differential equation
$$
\frac{dy}{dx} + \frac{x\,y}{x^2 - 1}
\;=\;
\frac{x^4 + 2x}{\sqrt{1 - x^2}}
$$
for $x \in (-1,1)$ is a first-order linear differential equation of the form
$$
\frac{dy}{dx} + P(x)\,y = Q(x).
$$
Step 2: Determine the Integrating Factor (IF)
The integrating factor is found by
$$
\text{IF} \;=\; e^{\int P(x)\,dx}.
$$
In our case, $P(x) = \frac{x}{x^2-1}$. Hence,
$$
\int P(x)\,dx
\;=\; \int \frac{x}{x^2 - 1}\,dx
\;=\; \frac{1}{2}\,\ln|x^2 - 1|.
$$
Therefore,
$$
\text{IF}
\;=\;
e^{\frac{1}{2}\ln|x^2 - 1|}
\;=\;
\sqrt{|x^2 - 1|}.
$$
Since $x \in (-1,1)$, we use $1 - x^2$ inside the square root (with appropriate sign):
$$
\text{IF}
\;=\;
\sqrt{1 - x^2}.
$$
Step 3: Multiply Through by the Integrating Factor
Multiplying the original differential equation by $\sqrt{1 - x^2}$ gives:
$$
y\,\sqrt{1 - x^2}
\;=\;
\int (x^4 + 2x)\,dx + C.
$$
Step 4: Integrate the Right-Hand Side
Compute
$$
\int (x^4 + 2x)\,dx
\;=\;
\int x^4\,dx
\;+\;
\int 2x\,dx
\;=\;
\frac{x^5}{5} \;+\; x^2.
$$
Thus the general solution becomes
$$
y\,\sqrt{1 - x^2}
\;=\;
\frac{x^5}{5} + x^2 + C.
$$
Step 5: Apply the Initial Condition
The solution curve passes through the origin, which implies $x=0 \implies y(0)=0.$ Plugging $x=0$ and $y=0$ into the equation:
$$
0 \cdot \sqrt{1 - 0^2}
\;=\;
\frac{0^5}{5} + 0^2 + C
\;\Longrightarrow\;
C = 0.
$$
Therefore, the particular solution is
$$
y
\;=\;
\frac{\frac{x^5}{5} + x^2}{\sqrt{1 - x^2}}
\;=\;
\frac{x^5 + 5x^2}{5\,\sqrt{1 - x^2}}.
$$
Step 6: Set Up the Required Definite Integral
We need to evaluate
$$
\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x)\,dx
\;=\;
\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}
\frac{x^5 + 5x^2}{5\,\sqrt{1 - x^2}}
\,dx.
$$
Observe that the $x^5$ term is an odd function, and integrating an odd function symmetrically about $0$ yields $0$. Hence only the $5x^2$ part contributes. This simplifies the integral to
$$
\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}
\frac{5x^2}{5\sqrt{1 - x^2}}
\,dx
\;=\;
\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}
\frac{x^2}{\sqrt{1 - x^2}}
\,dx.
$$
Because the integrand is even ($x^2/( \sqrt{1-x^2})$ is even), we can write
$$
\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}
\frac{x^2}{\sqrt{1 - x^2}}
\,dx
\;=\;
2 \int_{0}^{\frac{\sqrt{3}}{2}}
\frac{x^2}{\sqrt{1 - x^2}}
\,dx.
$$
Step 7: Use the Substitution $x = \sin\theta$
Let $x = \sin\theta$. Then $dx = \cos\theta\,d\theta$ and $\sqrt{1 - x^2} = \sqrt{1 - \sin^2\theta} = \cos\theta$. When $x=0$, $\theta = 0$; when $x = \frac{\sqrt{3}}{2}$, $\theta = \frac{\pi}{3}$. Thus the integral becomes
$$
2 \int_{0}^{\frac{\pi}{3}}
\frac{\sin^2\theta}{\cos\theta}
\cos\theta\,d\theta
\;=\;
2 \int_{0}^{\frac{\pi}{3}}
\sin^2\theta
\,d\theta.
$$
Step 8: Integrate $\sin^2\theta$
Recall that
$$
\sin^2\theta
=
\frac{1 - \cos(2\theta)}{2}.
$$
Therefore,
$$
2 \int_{0}^{\frac{\pi}{3}} \sin^2\theta\,d\theta
\;=\;
2 \int_{0}^{\frac{\pi}{3}} \frac{1 - \cos(2\theta)}{2}\,d\theta
\;=\;
\int_{0}^{\frac{\pi}{3}} \bigl(1 - \cos(2\theta)\bigr)\,d\theta.
$$
Hence,
$$
\int_{0}^{\frac{\pi}{3}}
\bigl(1 - \cos(2\theta)\bigr)\,d\theta
\;=\;
\left[\theta - \frac{\sin(2\theta)}{2}\right]_{0}^{\frac{\pi}{3}}.
$$
Step 9: Evaluate the Final Expression
At $\theta = \frac{\pi}{3}$:
$$
\theta
=
\frac{\pi}{3},
\quad
\sin(2\theta)
=
\sin\left(\frac{2\pi}{3}\right)
=
\frac{\sqrt{3}}{2}.
$$
Thus
$$
\left[\theta - \frac{\sin(2\theta)}{2}\right]_{0}^{\frac{\pi}{3}}
\;=\;
\left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right)
-
(0 - 0)
\;=\;
\frac{\pi}{3}
-
\frac{\sqrt{3}}{4}.
$$
Final Answer
The value of the given integral is
$$
\frac{\pi}{3}
\;-\;
\frac{\sqrt{3}}{4}.
$$
