# How do you integrate int sqrt(7+5x^2)/x using trig substitutions?

Jan 10, 2017

$I = \int \frac{\sqrt{7 + 5 {x}^{2}}}{x} \mathrm{dx}$

We will use the substitution $\sqrt{5} x = \sqrt{7} \tan \theta$. Differentiating this shows that $\sqrt{5} \mathrm{dx} = \sqrt{7} {\sec}^{2} \theta d \theta$. Then:

$I = \frac{1}{\sqrt{5}} \int \frac{\sqrt{7 + {\left(\sqrt{5} x\right)}^{2}}}{x} \sqrt{5} \mathrm{dx}$

Substituting in what we know (recall that $x = \frac{\sqrt{7}}{\sqrt{5}} \tan \theta$):

$I = \frac{1}{\sqrt{5}} \int \frac{\sqrt{7 + {\left(\sqrt{7} \tan \theta\right)}^{2}}}{\frac{\sqrt{7}}{\sqrt{5}} \tan \theta} \sqrt{7} {\sec}^{2} \theta d \theta$

Clearing up the square root, and moving around the constants:

$I = \frac{1}{\sqrt{5}} \frac{\sqrt{5}}{\sqrt{7}} \sqrt{7} \int \frac{\sqrt{7 + 7 {\tan}^{2} \theta}}{\tan} \theta {\sec}^{2} \theta d \theta$

All the constants outside the integral cancel. We can factor $\sqrt{7}$ from the square root:

$I = \int \frac{\sqrt{7} \sqrt{1 + {\tan}^{2} \theta}}{\tan} \theta {\sec}^{2} \theta d \theta$

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

$I = \sqrt{7} \int \frac{\sqrt{{\sec}^{2} \theta}}{\tan} \theta {\sec}^{2} \theta d \theta = \sqrt{7} \int {\sec}^{3} \frac{\theta}{\tan} \theta d \theta$

Writing as $\sin \theta$ and $\cos \theta$:

$I = \sqrt{7} \int \frac{1}{\cos} ^ 3 \theta \cos \frac{\theta}{\sin} \theta d \theta = \sqrt{7} \int \frac{1}{\cos} ^ 2 \theta \frac{1}{\sin} \theta d \theta = \sqrt{7} \int {\sec}^{2} \frac{\theta}{\sin} \theta d \theta$

Rewriting (again)!

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

Rearranging...

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

$I = \sqrt{7} \int \sin \frac{\theta}{\cos} \theta \frac{1}{\cos} \theta d \theta + \sqrt{7} \int \csc \theta d \theta$

$I = \sqrt{7} \int \tan \theta \sec \theta d \theta + \sqrt{7} \int \csc \theta d \theta$

These are common integrals:

$I = \sqrt{7} \sec \theta - \sqrt{7} \ln \left\mid \csc \theta + \cot \theta \right\mid + C$

We started with the substitution $\sqrt{5} x = \sqrt{7} \tan \theta$. This implies that $\tan \theta = \frac{\sqrt{5} x}{\sqrt{7}}$. In a right triangle, this means we have an opposite side of $\sqrt{5} x$ and an adjacent side of $\sqrt{7}$. Through the Pythagorean theorem, we see that the hypotenuse is $\sqrt{7 + 5 {x}^{2}}$.

Using our definitions of the trigonometric functions, we see that:

• $\sec \theta = \text{hypotenuse"/"adjacent} = \frac{\sqrt{7 + 5 {x}^{2}}}{\sqrt{7}}$
• $\csc \theta = \text{hypotenuse"/"opposite} = \frac{\sqrt{7 + 5 {x}^{2}}}{\sqrt{5} x}$
• $\cot \theta = \text{adjacent"/"opposite} = \frac{\sqrt{7}}{\sqrt{5} x}$

Then:

$I = \sqrt{7} \frac{\sqrt{7 + 5 {x}^{2}}}{\sqrt{7}} - \sqrt{7} \ln \left\mid \frac{\sqrt{7 + 5 {x}^{2}}}{\sqrt{5} x} + \frac{\sqrt{7}}{\sqrt{5} x} \right\mid + C$

$I = \sqrt{7 + 5 {x}^{2}} - \sqrt{7} \ln \left\mid \frac{\sqrt{7 + 5 {x}^{2}} + \sqrt{7}}{\sqrt{5} x} \right\mid + C$

Note that $\frac{1}{\sqrt{5}}$ can be pulled out of the logarithm through the rule $\log \left(\frac{a}{b}\right) = \log \left(a\right) - \log \left(b\right)$. It will be absorbed into the constant of integration, so this is most simply written:

$I = \sqrt{7 + 5 {x}^{2}} - \sqrt{7} \ln \left\mid \frac{\sqrt{7 + 5 {x}^{2}} + \sqrt{7}}{x} \right\mid + C$