# Integrating $\int\tan\theta\sec^5\theta\ d\theta$

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 $$\frac{(\sec\theta)^5}{5} + c$$

• $du = sec \theta tan \theta$ – Don Larynx Sep 23 '13 at 1:00
• $du = \sec\theta \tan\theta d\theta$. – Tunococ Sep 23 '13 at 1:00
• Wow, can't believe I missed that. Thank you! – ConfusingCalc Sep 23 '13 at 1:00
• You also happen to be off on your sentences by a word, as well. – Don Larynx Sep 23 '13 at 1:01
• Erg, I changed the title and accidentally removed one word too many. Haha knock a guy while he is down. – ConfusingCalc Sep 23 '13 at 1:03

$$\int \tan\theta\sec^5\theta\,d\theta = \int (\sec^4\theta)\Big( \tan\theta\sec\theta\, d\theta\Big) = \int u^4\, du.$$
• Interesting way to look at it. Are you suggesting $\int \frac{sinx}{cosx} * \frac{1}{cos^5(x)}$ then $\int sinx * (cos^5(x))^{-1}$? – ConfusingCalc Sep 23 '13 at 2:08
• How is it an overkill when OP substituted $u=\sec\theta$ and not an overkill when you choose to substitute $u=\sin\theta$? – peterwhy Sep 23 '13 at 2:08
• $\displaystyle\int \frac{1}{\cos^6 x}\Big(\sin x\,dx\Big)$ $\displaystyle=\int\frac{1}{u^6}\,(-du)$. ${}\qquad{}$ – Michael Hardy Sep 23 '13 at 2:16