Indefinite Integration
Trigonometric Substitution with Radical Denominator
nta_pyq_2023_apr
Grade 12
Question:
Let $f(x)=\displaystyle\int\dfrac{dx}{(3+4x^2)\sqrt{4-3x^2}}$, $|x|<\dfrac{2}{\sqrt{3}}$. If $f(0)=0$ and $f(1)=\dfrac{1}{\alpha}\tan^{-1}\!\left(\dfrac{\alpha}{\beta}\right)$, $\alpha,\beta>0$, then $\alpha^2+\beta^2$ is equal to _______.
Step-by-Step Solution
Key Concept: Substitute $x=\frac{1}{t}$, then $4t^2-3=\lambda^2$. The integral reduces to $-\frac{1}{3}\int\frac{d\lambda}{\lambda^2+\frac{25}{3}}=-\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{5}\tan^{-1}\frac{\sqrt{3}\lambda}{5}+C$.
$\alpha=5,\ \beta=\sqrt{3}$. $\alpha^2+\beta^2=28$.
Correct Answer: 28