This problem is relatively straight forward, but for some reason, my answer is off by the power of 1.
$$\int \tan \theta \sec^5\theta d\theta $$
The steps I take are
- Step 1. $$ u = \sec \theta $$ $$ du = \tan\theta $$
- Step 2. $$ \int u^5 du $$
- Step 3. $$ (u^6 / 6) $$
- Step 4. $$ \frac{(\sec\theta)^6}{6} + c $$
However, the answer according to wolfram is $$ \frac{(\sec\theta)^5}{5} + c $$
Best Answer
$$ \int \tan\theta\sec^5\theta\,d\theta = \int (\sec^4\theta)\Big( \tan\theta\sec\theta\, d\theta\Big) = \int u^4\, du. $$