# How do you integrate int 1/sqrt(-e^(2x) +49)dx using trigonometric substitution?

$\textcolor{b l u e}{\int \frac{1}{\sqrt{49 - {e}^{2 x}}} \mathrm{dx} = \frac{1}{7} \cdot \ln \left(7 - \sqrt{49 - {e}^{2 x}}\right) - \frac{1}{7} x + C}$

#### Explanation:

The given $\int \frac{1}{\sqrt{49 - {e}^{2 x}}} \mathrm{dx}$

Let ${e}^{x} = 7 \cdot \sin \theta$
and ${e}^{2 x} = 49 \cdot {\sin}^{2} \theta$
and ${e}^{x} \mathrm{dx} = 7 \cdot \cos \theta \cdot d \theta$
and $\mathrm{dx} = \frac{7 \cdot \cos \theta \cdot d \theta}{e} ^ x = \frac{7 \cdot \cos \theta \cdot d \theta}{7 \cdot \sin \theta} = \frac{\cos \theta \cdot d \theta}{\sin \theta}$

Let us do the integration

$\int \frac{1}{\sqrt{49 - {e}^{2 x}}} \mathrm{dx} = \int \frac{1}{\sqrt{49 - 49 \cdot {\sin}^{2} \theta}} \cdot \frac{\cos \theta \cdot d \theta}{\sin \theta}$

$\int \frac{1}{\sqrt{49 - {e}^{2 x}}} \mathrm{dx} = \int \frac{1}{\sqrt{49} \sqrt{1 - {\sin}^{2} \theta}} \cdot \frac{\cos \theta \cdot d \theta}{\sin \theta}$

$\int \frac{1}{\sqrt{49 - {e}^{2 x}}} \mathrm{dx} = \int \frac{1}{7 \cdot \sqrt{{\cos}^{2} \theta}} \cdot \frac{\cos \theta \cdot d \theta}{\sin \theta}$

$\int \frac{1}{\sqrt{49 - {e}^{2 x}}} \mathrm{dx} = \int \frac{1}{7 \cdot \cos \theta} \cdot \frac{\cos \theta \cdot d \theta}{\sin \theta}$

$\int \frac{1}{\sqrt{49 - {e}^{2 x}}} \mathrm{dx} = \int \frac{1}{7 \cdot \cancel{\cos} \theta} \cdot \frac{\cancel{\cos} \theta \cdot d \theta}{\sin \theta}$

$\int \frac{1}{\sqrt{49 - {e}^{2 x}}} \mathrm{dx} = \frac{1}{7} \cdot \int \frac{d \theta}{\sin \theta} = \frac{1}{7} \cdot \int \csc \theta \cdot d \theta$

$\int \frac{1}{\sqrt{49 - {e}^{2 x}}} \mathrm{dx} = \frac{1}{7} \cdot \ln \left(\csc \theta - \cot \theta\right) + C$

Now, we have to return it to the original variable x.

From our trigonometric substitution
${e}^{x} = 7 \cdot \sin \theta$

$\sin \theta = {e}^{x} / 7$ and our Right Triangle has opposite side$= {e}^{x}$ and hypotenuse$= 7$ and adjacent side$= \sqrt{49 - {e}^{2 x}}$

Therefore,

$\csc \theta = \frac{7}{e} ^ x$ and $\cot \theta = \frac{\sqrt{49 - {e}^{2 x}}}{e} ^ x$

And

$\int \frac{1}{\sqrt{49 - {e}^{2 x}}} \mathrm{dx} = \frac{1}{7} \cdot \ln \left(\csc \theta - \cot \theta\right) + C$

$\int \frac{1}{\sqrt{49 - {e}^{2 x}}} \mathrm{dx} = \frac{1}{7} \cdot \ln \left(\frac{7}{e} ^ x - \frac{\sqrt{49 - {e}^{2 x}}}{e} ^ x\right) + C$

$\int \frac{1}{\sqrt{49 - {e}^{2 x}}} \mathrm{dx} = \frac{1}{7} \cdot \ln \left(\frac{7 - \sqrt{49 - {e}^{2 x}}}{e} ^ x\right) + C$

$\int \frac{1}{\sqrt{49 - {e}^{2 x}}} \mathrm{dx} = \frac{1}{7} \cdot \left[\ln \left(7 - \sqrt{49 - {e}^{2 x}}\right) - \ln {e}^{x}\right] + C$

$\int \frac{1}{\sqrt{49 - {e}^{2 x}}} \mathrm{dx} = \frac{1}{7} \cdot \ln \left(7 - \sqrt{49 - {e}^{2 x}}\right) - \frac{1}{7} x + C$

God bless....I hope the explanation is useful.