Let f(x) = dx(3+4x^2)√4-3x^2, x

Question

Let $f(x) = \displaystyle\int\frac{dx}{(3+4x^2)\sqrt{4-3x^2}}$, $x<\dfrac2{\sqrt3}$. If $f(0)=0$ and $f(1) = \dfrac1{\alpha\beta}	an^{-1}\dfrac{\alpha}{\beta}$, $\alpha,\beta>0$, then $\alpha^2+\beta^2$ is equal to:

Answer: 28

Accepted Answer

Substitute $x=1/t$, then $4t^2-3=z^2$; after simplification the integral reduces to a standard $\dfrac{dz}{3z^2+25}$-type form:

$$f(x) = -\frac{1}{5\sqrt3}\tan^{-1}\left(\frac{\sqrt3}{5}\sqrt{4t^2-3}\right)+C,\qquad t=\frac1x$$

Using $f(0)=0$ fixes $C=\dfrac{\pi}{10\sqrt3}$. Evaluating at $x=1$:

$$f(1) = \frac1{5\sqrt3}\cot^{-1}\left(\frac{\sqrt3}5\right) = \frac1{5\sqrt3}\tan^{-1}\left(\frac{5}{\sqrt3}\right)$$

So $\alpha=5,\,\beta=\sqrt3$, giving $\alpha^2+\beta^2 = 25+3=28$.

Solutions · 1