# How do you integrate int sqrt(1-4x-2x^2) using trig substitutions?

Aug 20, 2016

$\int \left(\sqrt{1 - 4 x - 2 {x}^{2}}\right) \mathrm{dx} = 3 \frac{\sqrt{2}}{8} \left(2 \arcsin \left(\sqrt{\frac{2}{3}} \left(1 + x\right)\right) + \sin \left(2 \arcsin \left(\sqrt{\frac{2}{3}} \left(1 + x\right)\right)\right)\right) + C$

#### Explanation:

$\int \left(\sqrt{1 - 4 x - 2 {x}^{2}}\right) \mathrm{dx} = \sqrt{3} \int \left(\sqrt{1 - \frac{2}{3} {\left(x + 1\right)}^{2}}\right) \mathrm{dx}$

now calling $\sqrt{\frac{2}{3}} \left(x + 1\right) = \sin y$

$\sqrt{\frac{2}{3}} \mathrm{dx} = \cos y \mathrm{dy}$ and

$\int \left(\sqrt{1 - 4 x - 2 {x}^{2}}\right) \mathrm{dx} \equiv \sqrt{3} \sqrt{\frac{3}{2}} \int {\cos}^{2} y \mathrm{dy} = \frac{3}{\sqrt{2}} \left(\frac{y}{2} + \frac{1}{4} \sin \left(2 y\right)\right) + C$ and after substituting

$y = \arcsin \left(\sqrt{\frac{2}{3}} \left(x + 1\right)\right)$

$\int \left(\sqrt{1 - 4 x - 2 {x}^{2}}\right) \mathrm{dx} = 3 \frac{\sqrt{2}}{8} \left(2 \arcsin \left(\sqrt{\frac{2}{3}} \left(1 + x\right)\right) + \sin \left(2 \arcsin \left(\sqrt{\frac{2}{3}} \left(1 + x\right)\right)\right)\right) + C$