# How do you integrate int cosxsqrt(9+25sin^2x) using trig substitutions?

Sep 11, 2016

$\frac{5 \sin x \sqrt{9 + 25 {\sin}^{2} x} + 9 \ln \left\mid \sqrt{9 + 25 {\sin}^{2} x} + 5 \sin x \right\mid}{10} + C$

#### Explanation:

$\int \cos x \sqrt{9 + 25 {\sin}^{2} x} \mathrm{dx}$

First apply the non-trigonometric substitution $\sin x = u$ such that $\cos x \mathrm{dx} = \mathrm{du}$.

$= \int \sqrt{9 + 25 {u}^{2}} \mathrm{du}$

Now, we will apply the trigonometric substitution $u = \frac{3}{5} \tan \theta$. This implies that $\mathrm{du} = \frac{3}{5} {\sec}^{2} \theta d \theta$. These yield:

$= \int \sqrt{9 + 9 {\tan}^{2} \theta} \left(\frac{3}{5} {\sec}^{2} \theta d \theta\right)$

$= \frac{9}{5} \int \sqrt{1 + {\tan}^{2} \theta} \left({\sec}^{2} \theta d \theta\right)$

Recall that $1 + {\tan}^{2} \theta = {\sec}^{2} \theta$:

$= \frac{9}{5} \int {\sec}^{3} \theta d \theta$

The process of finding $\int {\sec}^{3} \theta d \theta$ is fairly complex, so click here to see how it's done.

$= \frac{9}{10} \left(\sec \theta \tan \theta + \ln \left\mid \sec \theta + \tan \theta \right\mid\right) + C$

Note that $\tan \theta = \frac{5 u}{3}$. This means we have a right triangle with an opposite side of $5 u$ and an adjacent side of $3$, resulting in a hypotenuse of $\sqrt{9 + 25 {u}^{2}}$.

The secant of this triangle is equal to the hypotenuse over the adjacent side, or $\frac{\sqrt{9 + 25 {u}^{2}}}{3}$. Replace these values in the solved integral to simplify:

$= \frac{9}{10} \left(\frac{\sqrt{9 + 25 {u}^{2}}}{3} \left(\frac{5 u}{3}\right) + \ln \left\mid \frac{\sqrt{9 + 25 {u}^{2}}}{3} + \frac{5 u}{3} \right\mid\right) + C$

$= \frac{5 u \sqrt{9 + 25 {u}^{2}}}{10} + \frac{9}{10} \ln \left\mid \frac{1}{3} \left(\sqrt{9 + 25 {u}^{2}} + 5 u\right) \right\mid + C$

Note that the $\frac{1}{3}$ in the logarithm can be taken out as a constant and be absorbed with $C$, the constant of integration.

$= \frac{5 u \sqrt{9 + 25 {u}^{2}} + 9 \ln \left\mid \sqrt{9 + 25 {u}^{2}} + 5 u \right\mid}{10} + C$

Since $u = \sin x$:

$= \frac{5 \sin x \sqrt{9 + 25 {\sin}^{2} x} + 9 \ln \left\mid \sqrt{9 + 25 {\sin}^{2} x} + 5 \sin x \right\mid}{10} + C$