# How do you integrate int x^2/sqrt(4x^2+25)dx using trigonometric substitution?

Mar 27, 2018

$\frac{1}{20} \left(\sqrt{4 x + 25}\right) x - \frac{5}{8} \ln \left(\frac{\sqrt{4 x + 25}}{5} + \frac{2}{5} x\right)$

#### Explanation:

$\int {x}^{2} / \sqrt{4 x + 25} \mathrm{dx}$

$L e t x = \frac{5}{2} \tan A , \mathrm{dx} = \left(\frac{5}{2} {\sec}^{2} a\right) \mathrm{dA}$

=intx^2/sqrt(4x+25)dx = int (5/2)^2(tan^2A sec^2A) /(sqrt(4(5/2)^2tan^2A+25)$\mathrm{dA}$

=intx^2/sqrt(4x+25)dx = int (5/2)^2(tan^2A sec^2A) /(sqrt(25tan^2A+25)$\mathrm{dA}$

$= \int {x}^{2} / \sqrt{4 x + 25} \mathrm{dx} = \int {\left(\frac{5}{2}\right)}^{2} \frac{{\tan}^{2} A {\sec}^{2} A}{5} \left(\sqrt{{\tan}^{2} A + 1}\right)$$\mathrm{dA}$

${\sec}^{2} A - 1 = {\tan}^{2} A$

$= \int \left(\frac{5}{4}\right) \frac{{\tan}^{2} A {\sec}^{2} A}{\sqrt{{\sec}^{2} A}}$$\mathrm{dA}$

$= \int \left(\frac{5}{4}\right) \frac{\left({\sec}^{2} A - 1\right) {\sec}^{2} A}{\sqrt{{\sec}^{2} A}}$$\mathrm{dA}$
$= \int \left(\frac{5}{4}\right) \left(\left({\sec}^{3} A - \sec A\right)\right)$$\mathrm{dA}$
$= \left(\frac{5}{4}\right) \left(\int \left({\sec}^{3} A\right) - \int \left(\sec A\right)\right)$$\mathrm{dA}$ -------- equation (1)

We need to find

$\int \left(\sec A\right)$ $\mathrm{dA}$ = $\int \sec A \frac{\sec A + \tan A}{\sec A + \tan A}$$\mathrm{dA}$

=$\int \sec A \frac{{\sec}^{2} A + \tan A \sec A}{\sec A + \tan A}$$\mathrm{dA}$

Let $B = \sec A + \tan A$
$\mathrm{dB}$$= \left(\sec A \tan A + {\sec}^{2} A\right)$ $\mathrm{dA}$
$= \int \frac{1}{B}$$\mathrm{dB}$
$= \ln | B | + c$
$= \ln | \sec A + \tan A | + c$ -----equation 2

We now need to find

$\int \left({\sec}^{3} A\right)$ $\mathrm{dA}$

Integrate by parts

$\int {\sec}^{2} A \sec A$ $\mathrm{dA}$
Let $C = \sec A$
$\mathrm{dC}$$= \sec A \tan A$$\mathrm{dA}$

$\mathrm{dD}$$= {\sec}^{2} A$ $\mathrm{dA}$
$D = \int {\sec}^{2} A$ $\mathrm{dA}$ $= \tan A + c o n s t$

Using by parts gives
$= \int {\sec}^{2} A \sec A$ $\mathrm{dA}$ $= C D - \int D$$\mathrm{dC}$/$\mathrm{dA}$
$= \sec A \tan A - \int \sec A \tan A \tan A$ $\mathrm{dA}$
$= \sec A \tan A - \int \left({\sec}^{2} A - 1\right) \sec A$ $\mathrm{dA}$
$= \sec A \tan A - \int \left({\sec}^{3} A - \sec A\right)$ $\mathrm{dA}$
$\int {\sec}^{3} A \sec A$ $\mathrm{dA}$$= \sec A \tan A - \left(\int {\sec}^{3} A - \int \sec A\right)$ $\mathrm{dA}$

substituting using equation 2
$2 \int {\sec}^{3} A$ $\mathrm{dA}$ =  secAtanA+ ln|secA+tanA)
 1/2(secAtanA+ ln|secA+tanA)) ----equation 3

$\int {x}^{2} / \sqrt{4 x + 25} \mathrm{dx}$ $= \left(\frac{5}{4}\right) \left(\int \left({\sec}^{3} A\right) - \int \left(\sec A\right)\right)$$\mathrm{dA}$

using the results from equations 2 and 3

$= \frac{5}{4} \left(\left(\frac{1}{2} \left(\sec A \tan A + \ln | \left(S e c A + \tan A\right) |\right)\right) - \ln | \left(S e c A + \tan A\right) |\right)$
=5/8(secAtanA + -ln|(SecA+tanA)|))

Now convert everything back to x
$x = \frac{5}{2} \tan A$, $\tan A = \frac{2}{5} x$, ${\tan}^{2} A = {\left(\frac{2}{5}\right)}^{2} {x}^{2}$
${\sec}^{2} A = {\tan}^{2} A + 1 = \frac{4 {x}^{2} + 25}{25}$
$\sec A = = \sqrt{\frac{4 {x}^{2} + 25}{25}} = \frac{\sqrt{4 x + 25}}{5}$

$\frac{1}{20} \left(\sqrt{4 x + 25}\right) x - \frac{5}{8} \ln \left(\frac{\sqrt{\left(4 x + 25\right)}}{5} + \frac{2}{5} x\right)$