Question

If $\begin{array}{l}\int_{0}^{\sqrt{3}}\frac{15x^3}{\sqrt{1+x^2 + \sqrt{(1+x^2)^3}}}dx = \alpha \sqrt{2}+\beta\sqrt{3},\end{array}$where $α, β$ are integers, then $α + β$ is equal to

Answer 1

Put

$\begin{array}{l}x = tan\theta \Rightarrow dx = sec^2\theta~ d\theta\end{array}$

$\begin{array}{l}\Rightarrow I = \int_{0}^{\frac{\pi}{3}}\frac{15\tan^3\theta . \sec^2\theta\ d\theta}{\sqrt{1+\tan^2\theta + \sqrt{\sec^6\theta}}}\end{array}$

$\begin{array}{l}\Rightarrow I = \int_{0}^{\frac{\pi}{3}}\frac{15\tan^2\theta \sec^2 \theta \ d\theta}{\sec \theta \sqrt{1+ \sec \theta}}\end{array}$

$\begin{array}{l}\Rightarrow I = \int_{0}^{\frac{\pi}{3}}\frac{15(\sec^2 \theta -1)\sec \theta \tan \theta \ d\theta}{\left(\sqrt{1+\sec \theta}\right)}\end{array}$

Now put 1 + secθ = t 2

$\begin{array}{l}\Rightarrow sec~\theta ~tan~\theta~ d\theta = 2tdt\end{array}$

$\begin{array}{l}\Rightarrow I = \int_{\sqrt{2}}^{\sqrt{3}}\frac{15((t^2-1)^2-1)2t \ dt}{t}\end{array}$

$\begin{array}{l}\Rightarrow I = 30\int_{\sqrt{2}}^{\sqrt{3}}\left(t^4-2t^2 + 1-1\right)dt\end{array}$

$\begin{array}{l}\Rightarrow I = 30\int_{\sqrt{2}}^{\sqrt{3}}\left(t^4-2t^2 + 1-1\right)dt\end{array}$

$\begin{array}{l}\Rightarrow I = 30 \left(\frac{t^5}{5}-\frac{2t^3}{3}\right)_{\sqrt{2}}^{\sqrt{3}}\end{array}$

$\begin{array}{l}=30\left[\left(\frac{9}{5}\sqrt{3}-2\sqrt{3}\right)-\left(\frac{4\sqrt{2}}{5}-\frac{4\sqrt{2}}{3}\right)\right]\end{array}$

$\begin{array}{l}=(54\sqrt{3}-60\sqrt{3})-(24\sqrt{2}-40\sqrt{2})\end{array}$

$\begin{array}{l}=16\sqrt{2}-6\sqrt{3}\end{array}$

$\begin{array}{l} \therefore \alpha = 16 ~\text{and}~ \beta = – 6\\\ \alpha + \beta = 10.\end{array}$