Question

\( \int_0^{1/2} \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} dx = \)

• A. \( \frac{1}{2} + \frac{\sqrt{3}}{2} \pi \)
• B. \( \frac{1}{2} - \frac{\sqrt{3}}{12} \pi \)
• C. \( -\frac{1}{2} + \frac{\sqrt{3}}{12} \pi \)
• D. \( \frac{1}{2} - \frac{\sqrt{3}}{2} \pi \)

Hint:

Use the substitution \( \sin^{-1} x = t \) to simplify the integral. This will also change the limits of integration. After substitution, the integral will involve \( t \sin t \), which can be solved using integration by parts. Remember the derivative of \( \sin^{-1} x \) and the trigonometric identities.

Answer 1

The Correct Option is B

Let \( I = \int_0^{1/2} \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} dx \).
Use substitution \( \sin^{-1} x = t \), so \( x = \sin t \).
Then \( dx = \cos t dt \).
When \( x = 0 \), \( t = \sin^{-1}(0) = 0 \).
When \( x = 1/2 \), \( t = \sin^{-1}(1/2) = \frac{\pi}{6} \).
The integral becomes: $$ I = \int_0^{\pi/6} \frac{\sin t \cdot t}{\sqrt{1 - \sin^2 t}} \cos t dt = \int_0^{\pi/6} \frac{t \sin t}{\sqrt{\cos^2 t}} \cos t dt $$ $$ = \int_0^{\pi/6} \frac{t \sin t}{|\cos t|} \cos t dt $$ In the interval \( [0, \pi/6] \), \( \cos t \ge 0 \), so \( |\cos t| = \cos t \).
$$ I = \int_0^{\pi/6} t \sin t dt $$ Use integration by parts: \( \int u dv = uv - \int v du \).
Let \( u = t \), \( dv = \sin t dt \).
Then \( du = dt \), \( v = -\cos t \).
$$ I = [-t \cos t]_0^{\pi/6} - \int_0^{\pi/6} (-\cos t) dt $$ $$ = \left( -\frac{\pi}{6} \cos\left(\frac{\pi}{6}\right) - (-0 \cdot \cos(0)) \right) + \int_0^{\pi/6} \cos t dt $$ $$ = -\frac{\pi}{6} \cdot \frac{\sqrt{3}}{2} + [\sin t]_0^{\pi/6} $$ $$ = -\frac{\sqrt{3} \pi}{12} + \left( \sin\left(\frac{\pi}{6}\right) - \sin(0) \right) $$ $$ = -\frac{\sqrt{3} \pi}{12} + \left( \frac{1}{2} - 0 \right) = \frac{1}{2} - \frac{\sqrt{3} \pi}{12} $$