# How do you find the indefinite integral of sqrt(25 + x^2)?

Mar 26, 2018

$\frac{1}{2} \left(x \sqrt{25 + {x}^{2}} + 25 \ln \left\mid x + \sqrt{25 + {x}^{2}} \right\mid\right) + C$

#### Explanation:

Use trigonometric substitution.

Draw a triangle with an angle $\theta$.

Label the opposite side as $x$ and adjacent side as $5$.

Now examine the triangle and notice that

$\tan \theta = \frac{x}{5}$

$\Rightarrow x = 5 \tan \theta$

$\Rightarrow \mathrm{dx} = 5 {\sec}^{2} \theta$ $d$$\theta$

By the Pythagorean Theorem, the hypotenuse of the triangle is:

$\sqrt{25 + {x}^{2}}$

So we can write

$\sec \theta = \frac{\sqrt{25 + {x}^{2}}}{5}$

$\Rightarrow \sqrt{25 + {x}^{2}} = 5 \sec \theta$

Let's now rewrite the integral in terms of $\theta$

intsqrt(25+x^2)dxrArrint(5sectheta)(5sec^2theta $d$theta)

$\Rightarrow 25 \int$ ${\sec}^{3} \theta$ $d$$\theta$

Integrating ${\sec}^{3} \theta$ can be tricky. There is a shortcut formula, but I'll do it the long way for teaching purposes.

Separate (sec^3theta $d$theta) into (sectheta*sec^2theta $d$theta) and use integration by parts.

Let:

$u = \sec \theta$
$\mathrm{du} = \sec \theta \tan \theta$ $d$$\theta$

$\mathrm{dv} = {\sec}^{2} \theta$ $d$$\theta$
$v = \tan \theta$

Then:

$\Rightarrow \int$ ${\sec}^{3} \theta$ $d$$\theta = \sec \theta \tan \theta - \int$ ${\tan}^{2} \theta \sec \theta$ $d$$\theta$

$\Rightarrow \int$ ${\sec}^{3} \theta$ $d$$\theta = \sec \theta \tan \theta - \int$ $\left({\sec}^{2} \theta - 1\right) \sec \theta$ $d$$\theta$

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

$\Rightarrow 2 \int$ ${\sec}^{3} \theta$ $d$$\theta = \sec \theta \tan \theta + \int$ $\sec \theta$ $d$$\theta$

$\Rightarrow 2 \int$ ${\sec}^{3} \theta$ $d$$\theta = \sec \theta \tan \theta + \ln \left\mid \sec \theta + \tan \theta \right\mid$

$\Rightarrow \int$ ${\sec}^{3} \theta$ $d$$\theta = \frac{1}{2} \left(\sec \theta \tan \theta + \ln \left\mid \sec \theta + \tan \theta \right\mid\right) + C$

And let's not forget that our integral was multiplied by 25!

$\Rightarrow 25 \int$ ${\sec}^{3} \theta$ $d$$\theta = \frac{25}{2} \left(\sec \theta \tan \theta + \ln \left\mid \sec \theta + \tan \theta \right\mid\right) + C$

Now put everything back in terms of $x$

$\Rightarrow \frac{25}{2} \left(\frac{\sqrt{25 + {x}^{2}}}{5} \frac{x}{5} + \ln \left\mid \frac{\sqrt{25 + {x}^{2}}}{5} + \frac{x}{5} \right\mid\right) + C$

$\Rightarrow \frac{1}{2} \left(x \sqrt{25 + {x}^{2}} + 25 \ln \left\mid \frac{\sqrt{25 + {x}^{2}} + x}{5} \right\mid\right) + C$

$\Rightarrow \frac{1}{2} \left(x \sqrt{25 + {x}^{2}} + 25 \ln \left\mid x + \sqrt{25 + {x}^{2}} \right\mid - 25 \ln 5\right) + C$

Then we can absorb the constant $- \frac{25 \ln 5}{2}$ into $C$ to get our final answer:

$\Rightarrow \frac{1}{2} \left(x \sqrt{25 + {x}^{2}} + 25 \ln \left\mid x + \sqrt{25 + {x}^{2}} \right\mid\right) + C$