<p>If \(I = \displaystyle\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \sqrt{\frac{1-x}{1+x}}\,\sin^{-1}x\,dx = \frac{\pi}{M} - \sqrt{N}\), find the value of \((M+N)\).</p>
Step-by-Step Solution
Key Concept: Recognize that the integrand f(x) = √[(1-x)/(1+x)]·sin⁻¹(x) is an odd function (product of even and odd functions), so the integral over a symmetric interval [-a,a] equals zero. This transforms the problem into finding M and N from the given form.
Step 1: Rewrite the integrand.
The integrand is $\sqrt{\frac{1-x}{1+x}}\,\sin^{-1}x$.
To simplify the square root term, multiply the numerator and denominator by $\sqrt{1-x}$:
$$ \sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{(1-x)(1-x)}{(1+x)(1-x)}} = \sqrt{\frac{(1-x)^2}{1-x^2}} = \frac{|1-x|}{\sqrt{1-x^2}} $$
For the given integration interval $[-\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}]$, we have $1-x > 0$, so $|1-x| = 1-x$.
Thus, the integrand becomes $\frac{1-x}{\sqrt{1-x^2}}\,\sin^{-1}x$.
The integral $I$ can be split into two parts:
$$ I = \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \left(\frac{1}{\sqrt{1-x^2}} - \frac{x}{\sqrt{1-x^2}}\right)\sin^{-1}x\,dx $$
$$ I = \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{\sin^{-1}x}{\sqrt{1-x^2}}\,dx - \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx $$
Step 2: Analyze the parity of the terms.
Let $f_1(x) = \frac{\sin^{-1}x}{\sqrt{1-x^2}}$.
$$ f_1(-x) = \frac{\sin^{-1}(-x)}{\sqrt{1-(-x)^2}} = \frac{-\sin^{-1}x}{\sqrt{1-x^2}} = -f_1(x) $$
Thus, $f_1(x)$ is an odd function. For an odd function integrated over a symmetric interval $[-a, a]$, the integral is zero:
$$ \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f_1(x)\,dx = 0 $$
Let $f_2(x) = \frac{x\sin^{-1}x}{\sqrt{1-x^2}}$.
$$ f_2(-x) = \frac{(-x)\sin^{-1}(-x)}{\sqrt{1-(-x)^2}} = \frac{(-x)(-\sin^{-1}x)}{\sqrt{1-x^2}} = \frac{x\sin^{-1}x}{\sqrt{1-x^2}} = f_2(x) $$
Thus, $f_2(x)$ is an even function. For an even function integrated over a symmetric interval $[-a, a]$, the integral is twice the integral from $0$ to $a$:
$$ \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f_2(x)\,dx = 2\int_0^{\frac{\sqrt{3}}{2}} f_2(x)\,dx $$
Therefore, the integral $I$ simplifies to:
$$ I = 0 - 2\int_0^{\frac{\sqrt{3}}{2}} \frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx $$
Step 3: Evaluate the definite integral.
Consider the integral $J = \int_0^{\frac{\sqrt{3}}{2}} \frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx$.
Let $u = \sin^{-1}x$. Then $du = \frac{1}{\sqrt{1-x^2}}\,dx$.
Also, $x = \sin u$.
The limits of integration change as follows:
When $x=0$, $u=\sin^{-1}(0)=0$.
When $x=\frac{\sqrt{3}}{2}$, $u=\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}$.
Substituting these into the integral $J$:
$$ J = \int_0^{\frac{\pi}{3}} u \sin u\,du $$
Now, integrate by parts using the formula $\int p\,dq = pq - \int q\,dp$.
Let $p=u$ and $dq=\sin u\,du$.
Then $dp=du$ and $q=-\cos u$.
$$ J = \left[-u\cos u\right]_0^{\frac{\pi}{3}} - \int_0^{\frac{\pi}{3}} (-\cos u)\,du $$
$$ J = \left[-u\cos u\right]_0^{\frac{\pi}{3}} + \int_0^{\frac{\pi}{3}} \cos u\,du $$
$$ J = \left[\left(-\frac{\pi}{3}\cos\frac{\pi}{3}\right) - (-0\cos 0)\right] + \left[\sin u\right]_0^{\frac{\pi}{3}} $$
$$ J = \left[-\frac{\pi}{3}\cdot\frac{1}{2} - 0\right] + \left[\sin\frac{\pi}{3} - \sin 0\right] $$
$$ J = -\frac{\pi}{6} + \frac{\sqrt{3}}{2} - 0 $$
$$ J = -\frac{\pi}{6} + \frac{\sqrt{3}}{2} $$
Step 4: Determine the value of $I$ and find $M$ and $N$.
From Step 2, $I = -2J$.
$$ I = -2\left(-\frac{\pi}{6} + \frac{\sqrt{3}}{2}\right) $$
$$ I = \frac{2\pi}{6} - 2\frac{\sqrt{3}}{2} $$
$$ I = \frac{\pi}{3} - \sqrt{3} $$
The problem states that $I = \frac{\pi}{M} - \sqrt{N}$.
Comparing our result with the given form:
$$ \frac{\pi}{3} - \sqrt{3} = \frac{\pi}{M} - \sqrt{N} $$
By inspection, $M=3$ and $N=3$.
Step 5: Calculate $(M+N)$.
$$ M+N = 3+3 = 6 $$
Correct Answer: 6