# How do you integrate int xsqrt(3 + x^2)dx using trigonometric substitution?

Jan 24, 2017

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

#### Explanation:

You do not really need a trigonometric substitution here. As:

$d \left(3 + {x}^{2}\right) = 2 x \mathrm{dx}$

you can just substitute:

$t = 3 + {x}^{2}$

and have:

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

However if you want to perform a trigonometric substitution, the right substitution when under the square root there is a sum of square is:

$x = \sqrt{3} \tan t$ with $t \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$

$\mathrm{dx} = \sqrt{3} \frac{\mathrm{dt}}{{\cos}^{2} t}$

In this case we have:

$\int x \sqrt{3 + {x}^{2}} \mathrm{dx} = 3 \int \tan t \sqrt{3 + 3 {\tan}^{2} t} \frac{\mathrm{dt}}{\cos} ^ 2 t = 3 \sqrt{3} \int \tan \frac{t}{\cos} ^ 2 t \sqrt{1 + {\tan}^{2} t} \mathrm{dt}$

we now use the trigonometric identity:

$1 + {\tan}^{2} t = \frac{1}{\cos} ^ 2 t$

and have:

$\int x \sqrt{3 + {x}^{2}} \mathrm{dx} = 3 \sqrt{3} \int \tan \frac{t}{\cos} ^ 2 t \sqrt{\frac{1}{\cos} ^ 2 t} \mathrm{dt}$

Then we note that $\cos t > 0$ in the interval $t \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$, so:

$\int x \sqrt{3 + {x}^{2}} \mathrm{dx} = 3 \sqrt{3} \int \tan \frac{t}{\cos} ^ 3 t \mathrm{dt} = 3 \sqrt{3} \int \sin \frac{t}{\cos} ^ 4 t \mathrm{dt} = - 3 \sqrt{3} \int \frac{\mathrm{dc} o s t}{\cos} ^ 4 t = \frac{\sqrt{3}}{\cos} ^ 3 t + C$

To substitute back $x$, we need to express $\cos t$ in terms of $\tan t$:

$\cos t = \sqrt{\frac{1}{1 + {\tan}^{2} t}} = \sqrt{\frac{1}{1 + {x}^{2} / 3}} = \frac{\sqrt{3}}{\sqrt{3 + {x}^{2}}}$

so that:

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