Evaluating $\int (\tan^3x+\tan^4x) dx $ using substitution solving $$\int (\tan^3x+\tan^4x) dx $$ using substitution $$t = \tan x$$
My approach has led me to $ \int (1+t)t\sin^2xdt$ which has an $x$ too much and isn't easily solvable for me. If I remove the $x$ I get $\sin^2(\arctan(t))$ and that's not too nice to work with...
 A: Let's wrap it up. Let $t = tanx$, so $dt = sec^2xdx = (1 + tan^2x)dx = (1 + t^2)dx$, and $dx = \dfrac{dt}{1 + t^2}$. So $I = \int\dfrac{t^3 + t^4}{1 + t^2} dt = \int \left(t - \dfrac{t}{1 + t^2} + t^2 - 1 + \dfrac{1}{1 + t^2}\right)dt = \dfrac{t^3}{3} + \dfrac{t^2}{2} - t - \dfrac{ln(1 + t^2)}{2} + tan^{-1}t + C = \dfrac{tan^3x}{3} + \dfrac{tan^2x}{2} - tanx - ln|secx| + x + C$
A: 
unfortunately $\sec$ is not used in my country

If you feel unconfortable with $\sec $ you may proceed from your last integral
\begin{equation*}
I=\int (1+t)t\sin ^{2}\left( \arctan t\right) \,dt
\end{equation*}
by using the following trigonometric identity
\begin{equation*}
\sin ^{2}x=\frac{\tan ^{2}x}{1+\tan ^{2}x},\qquad x=\arctan t,
\end{equation*}
which you can derive from the fundamental identity $\sin^2 x+\cos^2 x=1$ to obtain 
\begin{equation*}
\sin ^{2}\left( \arctan t\right) =\frac{t^{2}}{1+t^{2}}.
\end{equation*}
Consequently
\begin{equation*}
I=\int (1+t)t\frac{t^{2}}{1+t^{2}}\,dt=\int \frac{t^{3}+t^{4}}{1+t^{2}}\,dt,
\end{equation*}
which is integrable by partial fractions. By long division we compute 
\begin{equation*}
\frac{t^{4}+t^{3}}{1+t^{2}}=t^{2}+t-1+\frac{-t+1}{1+t^{2}}.
\end{equation*}
So
\begin{equation*}
I=\int \left( t^{2}+t-1\right) dt+\int \frac{-t+1}{1+t^{2}}dt.
\end{equation*}
Since 
\begin{eqnarray*}
\int \frac{-t+1}{1+t^{2}}dt &=&-\frac{1}{2}\int \frac{2t}{1+t^{2}}dt+\int 
\frac{1}{1+t^{2}}dt \\
&=&-\frac{1}{2}\ln \left( 1+t^{2}\right) +\arctan t+C,
\end{eqnarray*}
we thus have
\begin{eqnarray*}
I &=&\frac{1}{3}t^{3}+\frac{1}{2}t^{2}-t-\frac{1}{2}\ln \left(
1+t^{2}\right) +\arctan t+C \\
&=&\frac{1}{3}\tan ^{3}x+\frac{1}{2}\tan ^{2}x-\tan x-\frac{1}{2}\ln \left(
1+\tan ^{2}x\right) +x+C.
\end{eqnarray*}
A: Solve separately:
first: $\int \tan^3x dx$:
$$\int \tan^3x dx=\int \tan^2 x (\tan x dx)=\int (\sec^2 x-1)\tan x dx= $$
$$=\int \sec^2x \tan x dx -\int \tan xdx=\int (\sec x)^1( \sec x\tan x dx)+\int \frac{-\sin x dx}{\cos x}=$$
$$=\frac{\sec^2 x}{2}+\ln |\cos x| +c. $$
Second: $\int \tan^4 x dx$:
$$\int \tan^4 x dx=\int (\sec^2x-1)^2dx=\int (\sec^4 x-2\sec^2x +1)dx=  $$
$$=\int \sec^4 x dx -2\int \sec^2 x dx+\int dx = $$
$$=\int (1+\tan^2x)\sec^2x dx -2\int \sec^2x +x+c= $$
$$=\int \sec^2x dx+\int \tan^2x \sec^2 x dx -2\int \sec^2 xdx +x+c= $$
$$\int (\tan x)^2 \sec^2x dx -\int \sec^2 x dx+x+c =$$
$$=\frac{\tan^3 x}{3}-\tan x+x+c. $$
The, the answer is
$$\int(\tan^2x+\tan^4x)dx = \frac{1}{2}\sec^2x+\ln|\cos x|+\frac{1}{3}\tan^3x -\tan x+x+c.$$
Note: is easy to see that my solution and the LAcarguy above differ only a constant.
