Question

Find the antiderivative of ${\tan }^2\left(x\right)dx$.

Answer 1

Compute the antiderivative:

We can find the antiderivative as,

$\int {\tan }^2\left(x\right)dx=\int \frac{{\sin }^2\left(x\right)}{{\cos }^2\left(x\right)}dx$

$$\begin{gathered} =\int \frac{\left(1-{\cos }^2\left(x\right)\right)}{{\cos }^2\left(x\right)}dx\left[\because {\sin }^2\left(x\right)+{\cos }^2\left(x\right)=1 \\ \Rightarrow {\sin }^2\left(x\right)=1-{\cos }^2\left(x\right)\right] \end{gathered}$$

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

$=\int \left({sec}^2\left(x\right)-1\right)dx\left[\because \frac{1}{{\cos }^2\left(x\right)}={sec}^2\left(x\right)\right]$

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

$=\tan \left(x\right)-x+C\left[\because \int {sec}^2\left(x\right)dx=\tan \left(x\right),\int 1dx=x\right]$

Hence, the antiderivative of ${\tan }^2\left(x\right)dx$ is $\tan \left(x\right)-x+C$, where $C$ is constant of integration.