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  

\[\begin{array}{l} x = \tan \theta \Rightarrow dx = \sec^2 \theta \, d\theta \end{array}\]\[\Rightarrow I = \int_{0}^{\frac{\pi}{3}} \frac{15 \tan^3 \theta \cdot \sec^2 \theta \, d\theta}{\sqrt{1 + \tan^2 \theta + \sqrt{\sec^6 \theta}}}\]\[\Rightarrow I = \int_{0}^{\frac{\pi}{3}} \frac{15 \tan^2 \theta \sec^2 \theta \, d\theta}{\sec \theta \sqrt{1 + \sec \theta}}\]\[\Rightarrow I = \int_{0}^{\frac{\pi}{3}} \frac{15 (\sec^2 \theta - 1) \sec \theta \tan \theta \, d\theta}{\sqrt{1 + \sec \theta}}\]

Now put \(1 + \sec \theta = t^2\)

\[\Rightarrow \sec \theta \tan \theta \, d\theta = 2t \, dt\]\[\Rightarrow I = \int_{\sqrt{2}}^{\sqrt{3}} \frac{15 \left( (t^2 - 1)^2 - 1 \right) 2t \, dt}{t}\]\[\Rightarrow I = 30 \int_{\sqrt{2}}^{\sqrt{3}} \left( t^4 - 2t^2 + 1 - 1 \right) dt\]\[\Rightarrow I = 30 \int_{\sqrt{2}}^{\sqrt{3}} \left( t^4 - 2t^2 \right) dt\]\[\Rightarrow I = 30 \left( \frac{t^5}{5} - \frac{2t^3}{3} \right)_{\sqrt{2}}^{\sqrt{3}}\]\[= 30 \left[ \left( \frac{9}{5} \sqrt{3} - 2 \sqrt{3} \right) - \left( \frac{4 \sqrt{2}}{5} - \frac{4 \sqrt{2}}{3} \right) \right]\]\[= (54 \sqrt{3} - 60 \sqrt{3}) - (24 \sqrt{2} - 40 \sqrt{2})\]\[= 16 \sqrt{2} - 6 \sqrt{3}\]\[\therefore \alpha = 16 \text{ and } \beta = -6\]\[\alpha + \beta = 10\]