Question

integral of tan^2x

Answer 1

\[
\begin{split}
\int \tan^2(x) dx &= \int 1 - \sec^2(x) dx &\text{ note: } \tan^2(x)+1 = \sec^2(x) \\
&= (x + C_1) - \int \sec^2(x)dx &\\
&= x+C_1 - (\tan(x) + C_2) &\text{ note: } \frac{d}{dx} \tan(x) = \sec^2(x) \\
&= x - \tan(x) + C &\\
\end{split}
\]It's a tricky integral.
You just have to remember your trig identities and basic trig derivatives.

Answer 2

Whoops! Messed up a step \[
\begin{split}
\int \tan^2(x) dx &= \int \sec^2(x) -1dx &\text{ note: } \tan^2(x)+1 = \sec^2(x) \\
&= \int \sec^2(x)dx - \int 1dx &\\
&= \int \sec^2(x)dx - (x + C_1) &\\
&= (\tan(x) + C_2) - (x+C_1) &\text{ note: } \frac{d}{dx} \tan(x) = \sec^2(x) \\
&= \tan(x) - x+ C &\\
\end{split}
\]