# How do you integrate int xsqrt(x^2+4) by trigonometric substitution?

Apr 30, 2018

$\frac{8 {\left(\sec \theta\right)}^{3}}{3} + C$

#### Explanation:

Let x=$2 \tan \theta$
$\frac{\mathrm{dx}}{d \theta} = 2 {\left(\sec \theta\right)}^{2}$
$\mathrm{dx} = 2 {\left(\sec \theta\right)}^{2} \left(d \theta\right)$

$\int x \sqrt{{x}^{2} + 4} \mathrm{dx}$

=$\int 2 \tan \theta \sqrt{{\left(2 \tan \theta\right)}^{2} + 4} \times 2 {\left(\sec \theta\right)}^{2} \left(d \theta\right)$

=$\int 2 \tan \theta \sqrt{4 {\left(\tan \theta\right)}^{2} + 4} \times 2 {\left(\sec \theta\right)}^{2} d \theta$

=$\int 2 \tan \theta \times 2 \sec \theta \times 2 {\left(\sec \theta\right)}^{2} d \theta$

=$8 \int \tan \theta {\left(\sec \theta\right)}^{3} d \theta$

**

• Since $\frac{d}{\mathrm{dx}} \sec \theta = \sec \theta \tan \theta$

**

=$8 \int {\left(\sec \theta\right)}^{2} \times \left(\tan \theta \sec \theta\right)$

=$8 \times {\left(\sec \theta\right)}^{3} / 3 + C$

=$\frac{8 {\left(\sec \theta\right)}^{3}}{3} + C$