Indefinite Integration
Indefinite Integration
nta_pyq_2025_apr
Grade 12
Question:
If $\displaystyle\int e^x\!\left(\frac{x\sin^{-1}x}{\sqrt{1-x^2}} + \frac{\sin^{-1}x}{(1-x^2)^{3/2}} + \frac{x}{1-x^2}\right)dx = g(x)+C$, where $C$ is the constant of integration, then $g\!\left(\dfrac{1}{2}\right)$ equals:
$\dfrac{\pi}{4}\sqrt{\dfrac{e}{3}}$
$\dfrac{\pi}{6}\sqrt{\dfrac{e}{3}}$
$\dfrac{\pi}{4}\sqrt{\dfrac{e}{2}}$
$\dfrac{\pi}{6}\sqrt{\dfrac{e}{2}}$
Step-by-Step Solution
Key Concept: Recognise that $\dfrac{\sin^{-1}x}{(1-x^2)^{3/2}} + \dfrac{x}{1-x^2}$ is the derivative of $f(x) = \dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}$, so the integrand equals $e^x(f(x)+f'(x))$ and the result is $e^x f(x) + C$.
Let $f(x) = \dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}$. Differentiating:
$$f'(x) = \frac{\sin^{-1}x}{(1-x^2)^{3/2}} + \frac{x}{1-x^2}.$$
Hence the integrand is $e^x(f(x)+f'(x))$, so
$$g(x) = e^x\cdot f(x) = e^x\cdot\frac{x\sin^{-1}x}{\sqrt{1-x^2}}.$$
$$g\!\left(\frac{1}{2}\right) = e^{1/2}\cdot\frac{\frac{1}{2}\cdot\frac{\pi}{6}}{\sqrt{\frac{3}{4}}} = \sqrt{e}\cdot\frac{\pi/12}{\sqrt{3}/2} = \frac{\pi}{6}\sqrt{\frac{e}{3}}.$$
Correct Answer: 2