# Evaluate: int(1+2sinx/cos^2x)dx?

May 28, 2018

$I = \tan \left(x\right) + 2 \sec \left(x\right) + C$

#### Explanation:

I assume you want to integrate

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

Splitting into two terms

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

First integral
Remember color(blue)((tan(x))'=sec^2(x)

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

$\textcolor{w h i t e}{I} = \int {\sec}^{2} \left(x\right) \mathrm{dx}$

$\textcolor{w h i t e}{I} = \tan \left(x\right) + {C}_{1}$

Second integral
Make a substitution color(red)(u=cos(x)=>du=-sin(x)dx

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

color(white)(I_1)=-2int1/u^2du larr color(red)("The substitution"

$\textcolor{w h i t e}{{I}_{1}} = \frac{2}{u} + {C}_{2}$

$\textcolor{w h i t e}{{I}_{1}} = \frac{2}{\cos} \left(x\right) + {C}_{2}$

$\textcolor{w h i t e}{{I}_{1}} = 2 \sec \left(x\right) + {C}_{2}$

Combining these

$I = \tan \left(x\right) + 2 \sec \left(x\right) + C$