# How do you integrate int (sin3x)/sqrt(5+cos3x) using substitution?

Apr 10, 2018

$\int \frac{\sin 3 x}{\sqrt{5 + \cos 3 x}} \mathrm{dx} = - \frac{2}{3} \sqrt{5 + \cos 3 x} + C$

#### Explanation:

$I = \int \frac{\sin 3 x}{\sqrt{5 + \cos 3 x}} \mathrm{dx}$

Generally, when we have integrals where two trigonometric functions are being divided, we should tend to substitute them such that the numerator or denominator cancels out.

Firstly, let $u = 5 + \cos 3 x$. Then,

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[5 + \cos 3 x\right] = \frac{d}{\mathrm{dx}} \cos 3 x$

We have to use the $\textcolor{red}{\text{chain rule}}$, which states that:

$\left[f \left(g \left(x\right)\right)\right] ' = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

In our case, $f \left(x\right) = \cos x$ and $g \left(x\right) = 3 x$.

$\therefore \left[\cos 3 x\right] ' = - 3 \sin 3 x$

$\frac{d}{\mathrm{dx}} \cos 3 x = - 3 \sin 3 x$

We have:

$\mathrm{du} = - 3 \sin 3 x \mathrm{dx} \implies \mathrm{dx} = - \frac{1}{3 \sin 3 x} \mathrm{du}$

By substituting $u$ into the integral, we have:

$\int \frac{\sin 3 x}{\sqrt{5 + \cos 3 x}} \mathrm{dx} = \int \frac{\sin 3 x}{\sqrt{u}} \cdot \frac{\mathrm{du}}{- 3 \sin 3 x}$

You can see that the $\sin 3 x$ cancels, as we needed.

$\implies I = - \frac{1}{3} \int \frac{\cancel{\sin 3 x}}{\sqrt{u}} \cdot \frac{\mathrm{du}}{\cancel{\sin 3 x}} = - \frac{1}{3} \int \frac{1}{\sqrt{u}} \mathrm{du}$

Since $\frac{1}{\sqrt{u}} = {u}^{- \frac{1}{2}}$, we get:

$I = - \frac{1}{3} \int \frac{1}{\sqrt{u}} = - \frac{1}{3} \int {u}^{- \frac{1}{2}} \mathrm{du}$

We know that, for any $n \ne - 1$:

$\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C$

Thus, our integral is equal to:

$I = - \frac{1}{3} \cdot {u}^{1 - \frac{1}{2}} / \left(1 - \frac{1}{2}\right) + C = - \frac{2}{3} \sqrt{u} + C$

Knowing that $u = 5 + \cos 3 x$, we finally get our answer:

$I = - \frac{2}{3} \sqrt{5 + \cos 3 x} + C$