# How do you solve intxsqrt(x^2+8)*dx using trig substitution?

Apr 18, 2018

Let, $\sqrt{{x}^{2} + 8} = t \implies {x}^{2} + 8 = {t}^{2} \implies 2 x \mathrm{dx} = 2 t \mathrm{dt} \implies x \mathrm{dx} = t \mathrm{dt}$
$\therefore I = \int \sqrt{{x}^{2} + 8} \cdot x \mathrm{dx} = \int t \cdot t \mathrm{dt} = \int {t}^{2} \mathrm{dt} = {t}^{3} / 3 + c$
$\implies I = \frac{1}{3} {\left(t\right)}^{3} + c , w h e r e , t = \sqrt{{x}^{2} + 8}$
$= \frac{1}{3} {\left(\sqrt{{x}^{2} + 8}\right)}^{3} + c$

#### Explanation:

Decide which answer is better ?

Here,

$I = \int x \sqrt{{x}^{2} + 8} \mathrm{dx}$

Let, $x = 2 \sqrt{2} \tan u \implies \mathrm{dx} = 2 \sqrt{2} {\sec}^{2} u \mathrm{du}$

$\mathmr{and} \tan u = \frac{x}{2 \sqrt{2}} \implies {\tan}^{2} u = {x}^{2} / 8.$

$\therefore I = \int 2 \sqrt{2} \tan u \sqrt{8 {\tan}^{2} u + 8} \cdot 2 \sqrt{2} {\sec}^{2} u \mathrm{du}$

$= \int 2 \sqrt{2} \tan u 2 \sqrt{2} \sqrt{{\tan}^{2} u + 1} \cdot 2 \sqrt{2} {\sec}^{2} u \mathrm{du}$

$= 8 \int \tan u \sec u \cdot 2 \sqrt{2} {\sec}^{2} u \mathrm{du}$

$= 8 \left(2 \sqrt{2}\right) \int {\left[\sec u\right]}^{2} \left(\sec u \tan u\right) \mathrm{du}$

$= 16 \sqrt{2} \int {\left[\sec u\right]}^{2} \frac{d}{\mathrm{dx}} \left(\sec u\right) \mathrm{du}$

$= 16 \sqrt{2} \left({\left[\sec u\right]}^{3} / 3\right) + c$

$= \frac{16 \sqrt{2}}{3} {\left[\sqrt{{\tan}^{2} u + 1}\right]}^{3} + c , w h e r e , {\tan}^{2} u = {x}^{2} / 8$

$= \frac{16 \sqrt{2}}{3} {\left[\sqrt{{x}^{2} / 8 + 1}\right]}^{3} + c$

$= \frac{16 \sqrt{2}}{3} {\left(\frac{\sqrt{{x}^{2} + 8}}{2 \sqrt{2}}\right)}^{3} + c$

$= \frac{16 \sqrt{2}}{3} \times {\left(\sqrt{{x}^{2} + 8}\right)}^{3} / \left(16 \sqrt{2}\right) + c$

$= \frac{1}{3} {\left(\sqrt{{x}^{2} + 8}\right)}^{3} + c$