Question

If $\int \left[ 3t^2 \sin^{-1} \left( \frac{1}{-t \cos t} \right) \right] \, dt = f(t) \left( \sin^{-1} \left( \frac{1}{t} \right) \right) + c$, then $f(2) =$
Identify the correct option from the following:

• A. 2
• B. -12
• C. 8
• D. -16

Hint:

For integrals matching a specific form, hypothesize $f(t)$ based on the structure and verify by differentiation or direct integration.

Answer 1

The Correct Option is C

Step 1: Simplify the integrand
$\sin^{-1} \left( \frac{1}{-t \cos t} \right) = -\sin^{-1} \left( \frac{1}{t \cos t} \right)$. The integral is $\int 3t^2 \left( -\sin^{-1} \left( \frac{1}{t \cos t} \right) \right) \, dt$. Given the form, assume integration by parts or direct form matching. Step 2: Hypothesize $f(t)$
The result is $f(t) \sin^{-1} \left( \frac{1}{t} \right) + c$. Differentiate: $f'(t) \sin^{-1} \left( \frac{1}{t} \right) + f(t) \cdot \frac{-1}{t^2 \sqrt{1 - \frac{1}{t^2}}} = 3t^2 \left( -\sin^{-1} \left( \frac{1}{t \cos t} \right) \right)$. Test $f(t) = t^3$: $3t^2 \sin^{-1} \left( \frac{1}{t} \right) + t^3 \cdot \text{derivative term}$, adjust to match. Direct integration suggests $f(t) = t^3$. Step 3: Evaluate $f(2)$
If $f(t) = t^3$, $f(2) = 2^3 = 8$, matching option (3).