Integral $\int \tan^{5}(x)\text{ d}x$

Question

I would like guidance on evaluating $$\displaystyle\int \tan^{5}(x)\text{ d}x\text{.}$$ I have attempted using the Pythagorean identities to get $$\int\tan^{5}(x)\text{ d}x = \int \tan(x)\left[\sec^{2}(x)-1\right]^{2}\text{ d}x\text{.}$$ This doesn't look helpful. So I thought, why not turn only ONE of the $\tan^{2}$ terms to the $\sec^{2}-1$ form? This gives $$\int\tan^{5}(x)\text{ d}x = \int\tan^{3}(x)\sec^{2}(x)\text{ d}x-\int \tan^{3}(x)\text{ d}x\text{.}$$ Clearly the second term is $\dfrac{\tan^{4}(x)}{4}$ (ignoring the constant term for now). Using a similar trick, $$\begin{align} \int\tan^{3}(x)\text{ d}x &= \int \tan(x)\sec^{2}(x)\text{ d}x-\int\tan(x)\text{ d}x \\ &= \dfrac{\tan^{2}(x)}{2} - (-1)\ln|\cos(x)| \\ &= \dfrac{\tan^{2}(x)}{2}+\ln|\cos(x)|\text{.} \end{align}$$ So this suggests to me that $$\int\tan^{5}(x)\text{ d}x = \dfrac{\tan^{4}(x)}{4} - \dfrac{\tan^{2}(x)}{2}-\ln|\cos(x)| + C\text{.}$$ But the answer in Stewart (section 7.2., #31) is $$\dfrac{1}{4}\sec^{4}(x)-\tan^{2}(x)+\ln|\sec(x)|+C\text{.}$$ It's very clear where the $\ln|\sec(x)|$ term is coming from - and I tried to take the difference of my answer and Stewart's answer using Wolfram Alpha and unfortunately, the difference is not a constant.

Where did I go wrong?

Answer 1

Short answer: the answers are equivalent, you have nothing to worry about. :)

Long answer: \begin{align} \frac{\tan^4 x}{4} - \frac{\tan^2 x}{2} &= \frac{\tan^4 x + 2\tan^2 x}{4} - \tan^2 x\\ &= \frac{\sec^4 x- 1}{4} - \tan^2 x\\ &= \frac{\sec^4x}{4} - \tan^2 x + C \end{align}

Answer 2

Your answer is correct. We have $\tan^2x=\sec^2x-1$ so $\tan^4x=\sec^4x-2\sec^2x+1$. Replacing your $\tan^4x$ accordingly gives $\frac14\sec^4x-\frac12\sec^2x+\frac14.$ Replacing that $\sec^2x$ with $\tan^2x+1$ results in Stewart's solution. The answers do differ by some constant though.

Answer 3

$$z=\tan{x}\Rightarrow dz=(z^{2}+1)dx\\ z^{5}=(z^{2}+1)(z^{3}-z)+z\\ \displaystyle \int z^{5}dx=\int(z^{3}-z)dz+\int zdx=\frac{z^{4}}{4}-\frac{z^{2}}{2}+\int zdx \\ \displaystyle \int \tan^{5}{x}dx=\frac{\tan^{4}{x}}{4}-\frac{\tan^{2}{x}}{2}-\ln\left| \cos{x} \right|+c$$

Answer 4

There is an easy way to evaluate $\int \sin^n(x)\cos^m(x)\,dx$ when at least one of $n,m$ is odd. Say $n$ is odd. Then keep one factor of $\sin$ and convert all the rest to $\cos$, and substitute $u=\cos x, du=-\sin x\,dx$.

Here is this case ($n=5,m=-5$): \begin{align} \int\tan^5 x\,dx &= \int \frac{\sin^5 x}{\cos^5 x}\,dx = \int\frac{(\sin x)(\sin^2 x)^2}{\cos^5 x}\,dx \\ &= \int\frac{(\sin x)(1-\cos^2 x)^2}{\cos^5 x}\,dx = -\int \frac{(1-u^2)^2}{u^5}\,dx \\ &= -\int\frac{du}{u}+2\int\frac{du}{u^3}-\int\frac{du}{u^5} = -\log(u) -\frac{1}{u^2}+\frac{1}{4u^4}+C \\ &= -\log(\cos x) - \frac{1}{\cos^2 x}+\frac{1}{4\cos^4 x}+C \end{align}