Evaluating $\int (\tan^3x+\tan^4x) dx $ using substitution

Question

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...

Answer 1

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*}

Answer 2

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 $I = \int\dfrac{t^3 + t^4}{1 + t^2} dt = \int \left(t - \dfrac{t}{1 + t^2} + t^2 - 1 + \dfrac{1}{1 + t^2}\right)dt = \dfrac{t^3}{3} + \dfrac{t^2}{2} - t - \dfrac{ln(1 + t^2)}{2} + tan^{-1}t + C = \dfrac{tan^3x}{3} + \dfrac{tan^2x}{2} - tanx - ln|secx| + x + C$

Answer 3

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.