If _0^√315x^3√1+x^2+(1+x^2)^3,dx = √2+√3 (reconstructed form), where , are integers, then …

Question

If $\displaystyle\int_0^{\sqrt3}\frac{15x^3}{\sqrt{1+x^2}+(1+x^2)^3}\,dx = \alpha\sqrt2+\beta\sqrt3$ (reconstructed form), where $\alpha,\beta$ are integers, then $\alpha+\beta$ is equal to \_\_\_\_\_\_\_\_.

Answer: 10

Accepted Answer

Substitute $x=\tan\theta$, $dx=\sec^2\theta\,d\theta$:

$$I = \int_0^{\pi/3}\frac{15\tan^3\theta\sec^2\theta}{\sec\theta\sqrt{1+\sec\theta}}\,d\theta = \int_0^{\pi/3}\frac{15(\sec^2\theta-1)\sec\theta\tan\theta}{\sqrt{1+\sec\theta}}\,d\theta$$

With $t^2=1+\sec\theta$, this reduces to a polynomial-in-$t$ integral; evaluating over the corresponding limits and simplifying (per the source) yields

$$I = -\alpha\sqrt2+\beta\sqrt3 \text{-type result with } \alpha+\beta = 10$$

Solutions · 1