# How do you integrate int sqrt(1-7x^2) using trig substitutions?

Apr 2, 2017

$\int \setminus \sqrt{1 - 7 {x}^{2}} \setminus \mathrm{dx} = \frac{1}{2 \sqrt{7}} \arcsin \left(\sqrt{7} x\right) + \frac{1}{2} x \sqrt{1 - 7 {x}^{2}} + C$

#### Explanation:

Let:

$I = \int \setminus \sqrt{1 - 7 {x}^{2}} \setminus \mathrm{dx}$

We would aim to get an expression involving ${a}^{2} - {x}^{2}$ prior to substitution so we can rewrite as:

$I = \int \setminus \sqrt{1 - {\left(\sqrt{7} x\right)}^{2}} \setminus \mathrm{dx}$

If we compare to the trig identity $1 - {\sin}^{2} \theta \equiv {\cos}^{2} \theta$ then it may be worth trying the substitution:

$\sin \theta = \sqrt{7} x \implies \theta = \arcsin \left(\sqrt{7} x\right)$
and, $\cos \theta \frac{d \theta}{\mathrm{dx}} = \sqrt{7}$

Substituting into our integral gives:

$I = \int \setminus \sqrt{1 - {\sin}^{2} \theta} \setminus \left(\cos \frac{\theta}{\sqrt{7}}\right) \setminus d \theta$
$\setminus \setminus = \frac{1}{\sqrt{7}} \setminus \int \setminus \sqrt{{\cos}^{2} \theta} \cos \theta \setminus d \theta$
$\setminus \setminus = \frac{1}{\sqrt{7}} \setminus \int \setminus {\cos}^{2} \theta \setminus d \theta$

Now we use the identity:

$\cos 2 \theta \equiv {\cos}^{2} \theta - {\sin}^{2} \theta$
$\text{ } = {\cos}^{2} \theta - \left(1 - {\cos}^{2}\right) \theta$
$\text{ } = 2 {\cos}^{2} \theta - 1$

Which gives us:

$I = \frac{1}{\sqrt{7}} \setminus \int \setminus {\cos}^{2} \theta \setminus d \theta$
$\setminus \setminus = \frac{1}{\sqrt{7}} \setminus \int \setminus \frac{1}{2} \left(1 + \cos 2 \theta\right) \setminus d \theta$
$\setminus \setminus = \frac{1}{2 \sqrt{7}} \setminus \int \setminus 1 + \cos 2 \theta \setminus d \theta$
$\setminus \setminus = \frac{1}{2 \sqrt{7}} \left\{\theta + \frac{1}{2} \sin 2 \theta\right\} + C$
$\setminus \setminus = \frac{1}{2 \sqrt{7}} \theta + \frac{1}{4 \sqrt{7}} \sin 2 \theta + C$
$\setminus \setminus = \frac{1}{2 \sqrt{7}} \theta + \frac{1}{4 \sqrt{7}} \left(2 \sin \theta \cos \theta\right) + C$

And if we use:

$\cos \theta = \sqrt{1 - {\sin}^{2} \theta}$

then we can restore the substitution and get:

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