# Question #62823

Jan 7, 2018

$= \frac{1}{2 \sqrt{7}} {\sin}^{- 1} \left(\sqrt{7} w\right) + \frac{w}{2} \sqrt{1 - 7 {w}^{2}} + C$

#### Explanation:

$\int \sqrt{1 - 7 {w}^{2}} \mathrm{dw}$

Consider the substitution:

$\sqrt{7} w = \sin \theta$

$\to \mathrm{dw} = \frac{1}{\sqrt{7}} \cos \theta d \theta$

Substituting these into the integral:

$\int \sqrt{1 - {\sin}^{2} \theta} \cdot \frac{1}{\sqrt{7}} \cos \theta d \theta$

We can use $1 - {\sin}^{2} \theta = {\cos}^{2} \theta$ to get:

$= \frac{1}{\sqrt{7}} \int \sqrt{{\cos}^{2} \theta} \cdot \cos \theta d \theta$

$= \frac{1}{\sqrt{7}} \int \cos \theta \cos \theta d \theta = \frac{1}{\sqrt{7}} \int {\cos}^{2} \theta d \theta$

Now use the trig identity:

${\cos}^{2} \theta = \frac{1}{2} + \frac{1}{2} \cos 2 \theta$

So the integral now becomes:

$\frac{1}{\sqrt{7}} \int \frac{1}{2} + \frac{1}{2} \cos 2 \theta d \theta = \frac{1}{2 \sqrt{7}} \int 1 + \cos 2 \theta d \theta$

$= \frac{1}{2 \sqrt{7}} \left(\theta + \frac{1}{2} \sin 2 \theta\right) + C$

Now use the trig identity:

$\sin \left(2 \theta\right) = 2 \cos \theta \sin \theta$ to get:

$= \frac{1}{2 \sqrt{7}} \left(\theta + \cos \theta \sin \theta\right) + C$

$= \frac{1}{2 \sqrt{7}} \left(\theta + \sqrt{1 - {\sin}^{2} \theta} \sin \theta\right) + C$

Now we can reverse the substitution we started with:

$= \frac{1}{2 \sqrt{7}} \left({\sin}^{- 1} \left(\sqrt{7} w\right) + \sqrt{1 - 7 {w}^{2}} \sqrt{7} w\right) + C$

$= \frac{1}{2 \sqrt{7}} {\sin}^{- 1} \left(\sqrt{7} w\right) + \frac{w}{2} \sqrt{1 - 7 {w}^{2}} + C$