How to solve $ \int \tan{\theta}\sec^4{\theta}d\theta $ I got the answer of:
$$\large{\frac {\tan^2{\theta}}{2} + \frac {\tan^4{\theta}}{4}  + C}$$
Wolfram Alpha got: $\large{\frac {sec^4{\theta}}{4}}$.
I don't know what I did wrong: I set $\large{u = tan{\theta}}$.
This is problem 20 from the chapter 8 review in Ron Larson's 8th edition of Calculus.
 A: make a change of variable $u = \sec \theta, \quad du = \sec\theta \tan \theta \, d\theta$ with that $$\int \sec^4 \theta \tan \theta \, d\theta = \int u^3 \, du = \frac 14 u^4 + c$$
A: Note that $\sec^2 \theta= 1+\tan^2\theta$ so that $\sec^4\theta=1+2\tan^2\theta+\tan^4\theta$
So $$\frac {\sec^4\theta}4 +C=\frac {\tan^4\theta}4+\frac {\tan^2\theta}2+(C+\frac14)$$The two answers are compatible.
A: if you set $u = \sec{\theta}$ then you have $du = \tan{\theta}\sec{\theta}$  and so $$d\theta = \frac{du}{\tan{\theta}\sec{\theta}} =\frac{du}{u\tan{\theta}}$$ and so now your integral becomes $$\int{\tan{\theta}u^4 \times \frac{du}{utan{\theta}}}$${then you will get rid of you the $\tan{\theta}$ in your integral and you will cancel one $u$. So now you have $\int{u^3}du$ and easily you do the integration to get $$\frac{u^4}{4} + C$$
Now you substitute back your $u = \sec{\theta}$ and your final answer becomes $$\frac{\sec^4{\theta}}{4} + C^{\prime}$$
A: Approach by your substitution:
$$u=\tan{\theta}, du=\sec^2{\theta}d\theta$$
\begin{align}
\int{\tan{\theta}}\sec^4{\theta}d\theta&=\int{u\sec^2{\theta}}du\\&=\int{u(1+u^2)}du\\&=\int{u+u^3}du\\&=\frac12u^2+\frac14u^4+C\\&=\frac12\tan^2{\theta}+\frac14\tan^4{\theta}+C
\end{align}
Wolfram Alpha's Approach:
$$u=\sec{\theta}, du=\sec{\theta}\tan{\theta}d\theta$$
$$\int \sec^4 \theta \tan \theta \, d\theta = \int u^3 \, du = \frac 14 u^4 + c=\frac14\sec^4{\theta}+C_0$$
