# How do you integrate this?

## $\int \frac{1}{s e n x + 5}$ dx Is it possible to solve this integrate using another way not this way --> substituting u=$\tan \left(\frac{x}{2}\right)$

Apr 5, 2018

$I = \frac{\arctan \left(\cos \frac{x}{2 \sqrt{6}}\right) + \arctan \left(\frac{2 \sqrt{6} \tan \left(x\right)}{5}\right)}{2 \sqrt{6}} + C$

#### Explanation:

Yes, it's a little more complicated however, we seek

$I = \int \frac{1}{\sin \left(x\right) + 5} \mathrm{dx}$

Multiply the DEN and NUM by the conjugate color(green)(sin(x)-5

$I = \int \frac{\sin \left(x\right) - 5}{\left(\sin \left(x\right) + 5\right) \left(\sin \left(x\right) - 5\right)} \mathrm{dx}$

$\textcolor{w h i t e}{I} = \int \frac{\sin \left(x\right) - 5}{\left({\sin}^{2} \left(x\right) - 25\right)} \mathrm{dx}$

$\textcolor{w h i t e}{I} = - \int \frac{\sin \left(x\right)}{\left({\cos}^{2} \left(x\right) + 24\right)} \mathrm{dx} + \int \frac{5}{\left({\cos}^{2} \left(x\right) + 24\right)} \mathrm{dx}$

First integral:

${I}_{1} = - \int \frac{\sin \left(x\right)}{\left({\cos}^{2} \left(x\right) + 24\right)} \mathrm{dx}$

Make a substitution $u = \cos \left(x\right) \implies \mathrm{du} = - \sin \left(x\right) \mathrm{dx}$

${I}_{1} = \int \frac{1}{\left({u}^{2} + 24\right)} \mathrm{du}$

$\textcolor{w h i t e}{{I}_{1}} = \frac{1}{24} \int \frac{1}{{\left(\frac{u}{\sqrt{24}}\right)}^{2} + 1} \mathrm{du}$

$\textcolor{w h i t e}{{I}_{1}} = \arctan \frac{\cos \frac{x}{\sqrt{24}}}{\sqrt{24}} + C$

Second integral:

${I}_{2} = \int \frac{5}{\left({\cos}^{2} \left(x\right) + 24\right)} \mathrm{dx}$

$\textcolor{w h i t e}{{I}_{2}} = 5 \int \frac{{\sec}^{2} \left(x\right)}{\left(1 + 24 {\sec}^{2} \left(x\right)\right)} \mathrm{dx}$

$\textcolor{w h i t e}{{I}_{2}} = 5 \int \frac{{\sec}^{2} \left(x\right)}{24 {\tan}^{2} \left(x\right) + 25} \mathrm{dx}$

Make a substitution $u = \tan \left(x\right) \implies \mathrm{du} = {\sec}^{2} \left(x\right) \mathrm{dx}$

${I}_{2} = 5 \int \frac{1}{24 {u}^{2} + 25} \mathrm{du}$

$\textcolor{w h i t e}{{I}_{2}} = \frac{1}{5} \int \frac{1}{{\left(\frac{\sqrt{24} u}{5}\right)}^{2} + 1} \mathrm{du}$

$\textcolor{w h i t e}{{I}_{2}} = \arctan \frac{\frac{\sqrt{24} \tan \left(x\right)}{5}}{\sqrt{24}} + C$

Thus

$I = \frac{\arctan \left(\cos \frac{x}{\sqrt{24}}\right) + \arctan \left(\frac{\sqrt{24} \tan \left(x\right)}{5}\right)}{\sqrt{24}} + C$

$\textcolor{w h i t e}{I} = \frac{\arctan \left(\cos \frac{x}{2 \sqrt{6}}\right) + \arctan \left(\frac{2 \sqrt{6} \tan \left(x\right)}{5}\right)}{2 \sqrt{6}} + C$