How to find the following integral? $\int\tfrac{\sec^2(x)}{3+ 2 \tan(x)}\,\mathrm dx $

Question

Please show me how to find the following integral: $$\int\dfrac{\sec^2(x)}{3+ 2 \tan(x)}\,\mathrm dx $$ I think the solution will be something about the natural logarithm right ?

Answer 1

Set $$u = 3+2\tan(x)$$ then $$du = 2\sec^2(x)dx$$ this would then imply $$\int\dfrac{\sec^2(x)}{3+ 2 \tan(x)}\,\mathrm dx = \int\frac{\sec^2(x)dx}{u} = \int\frac{\frac{1}{2}du}{u} = \frac12\int\frac{du}{u} = \frac12\ln(u) +c $$
$$ = \frac12 \ln(3+2\tan(x)) +c $$ etc..

Answer 2

Recognize that $\frac{d}{dx} \tan x = \sec^2 x$, and then substitute $u = \tan x$, $du = \sec^2 x\, dx$.

Using that $\int \frac{dx}{x} = \ln |x|+C$, you next obtain: $$\int \frac{du}{3+2 u} = \frac{1}{2} \ln|3+2u|+C= \frac{1}{2} \ln|3+2 \tan x|+C.$$

Answer 3

$$\int\dfrac{\color{red}{\sec^2(x)}}{\color{blue}{3+ 2 \tan(x)}}\,dx$$ This is actually an easy u-substitution problem. Look at the denominator and the numerator. Note that:

$$\frac{d}{dx}(\color{blue}{3+ 2 \tan(x)})=2\color{red}{\sec^2 (x)}$$

Cool, and then the $2$ is easy to get rid of because it's just a constant. Although this is easy to solve in your mind, let's do u-substituion. We'll set the denominator to be $u$:

$$\begin{align} u&=\color{blue}{3+2\tan(x)} \\du&=2\color{red}{\sec^2(x)}\; dx \\&\dots \\ \color{red}{\sec^2x} &=\frac{du}{2dx} \\ \color{blue}{3+2\tan(x)} &= u \end{align}$$

Substitute: $$\require{cancel} \int\dfrac{\color{red}{\sec^2(x)}}{\color{blue}{3+ 2 \tan(x)}}\,dx= \int \dfrac{\frac{du}{2\cancel{dx}}}{u}\;\cancel{dx}=\int \dfrac{du}{2u}=\frac12 \int \frac{du}{u}$$

What is this? Well, remember this:

$$\frac{d}{dx} \ln u = \frac1{u} \frac{du}{dx}$$

So the integral of $\frac{du}{u}$ is very simply $\ln u$.

Our answer is therefore:

$$\frac12 \int \frac{du}{u}=\frac12\ln \left|\underbrace{u}_{\color{blue}{3+2\tan(x)}} \right|=\begin{equation}\boxed{\;\therefore \frac12 \ln \left|3+2\tan(x)\right|+C \;}\end{equation}$$