Question

Evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.) 3 tan2(x) sec(x) dx

Answer 1

Answer:

$\frac{1}{2}\text{sec}(x)\text{tan}(x)-\frac{1}{2}\text{ln}|\text{tan}(x)+\text{sec}(x)|+C$

Step-by-step explanation:

We have been given an integral $\int \:\text{tan}^2\left(x\right)\text{sec}\left(x\right)dx$. We are asked to evaluate the given integral.

We will use identity $\text{sec}^2(x)-\text{tan}^2(x)=1$ as $\text{tan}^2(x)=-1+\text{sec}^2(x)$:

$\int \:(-1+\text{sec}^2(x))\text{sec}\left(x\right)dx$

$\int \:-\text{sec}(x)+\text{sec}^3(x)dx$

$\int \:-\text{sec}(x)dx+\int \:\text{sec}^3(x)dx$

Using common integral $\int \:-\text{sec}(x)dx=\text{ln}|\text{tan}(x)+\text{sec}(x)|$, we will get:

$-\text{ln}|\text{tan}(x)+\text{sec}(x)|+\int \:sec^3(x)dx$

Using integral reduction $\int \:\text{sec}^n(x)dx=\frac{\text{sec}^{n-1}(x)\text{sin}(x)}{n-1}+\frac{n-2}{n-1}\int \: \text{sec}^{n-2}(x)dx$, we will get:

$-\text{ln}|\text{tan}(x)+\text{sec}(x)|+\frac{\text{sec}^{3-1}(x)\text{sin}(x)}{3-1}+\frac{3-2}{3-1}\int \: \text{sec}^{3-2}(x)dx$

$-\text{ln}|\text{tan}(x)+\text{sec}(x)|+\frac{\text{sec}^{2}(x)\text{sin}(x)}{2}+\frac{1}{2}\int \: \text{sec}(x)dx$      

$-\text{ln}|\text{tan}(x)+\text{sec}(x)|+\frac{\text{sec}(x)\frac{\text{sin}(x)}{\text{cos}(x)}}{2}+\frac{1}{2}\int \: \text{sec}(x)dx$

$-\text{ln}|\text{tan}(x)+\text{sec}(x)|+\frac{1}{2}(\text{sec}(x)\text{tan}(x))+\frac{1}{2}(\text{ln}|\text{tan}(x)+\text{sec}(x)|)+C$

$\frac{1}{2}\text{sec}(x)\text{tan}(x)-\frac{1}{2}\text{ln}|\text{tan}(x)+\text{sec}(x)|+C$

Therefore, our required integral would be $\frac{1}{2}\text{sec}(x)\text{tan}(x)-\frac{1}{2}\text{ln}|\text{tan}(x)+\text{sec}(x)|+C$.