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

user66733
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ConfusingCalc
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2 Answers2

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$$ \int \tan\theta\sec^5\theta\,d\theta = \int (\sec^4\theta)\Big( \tan\theta\sec\theta\, d\theta\Big) = \int u^4\, du. $$

Michael Hardy
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You can also write the integral in terms of sines and cosines. You will then see that it is essentially a function in terms of cosines, since the sine is accounted for by the chain rule. A U-sub is overkill here. Bring the cosine term "up" and apply power rule for integrating.

imranfat
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  • 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 02:08
  • 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 02:08
  • $\displaystyle\int \frac{1}{\cos^6 x}\Big(\sin x,dx\Big)$ $\displaystyle=\int\frac{1}{u^6},(-du)$. ${}\qquad{}$ – Michael Hardy Sep 23 '13 at 02:16
  • Peter, as Michael indicated, I am not using a substitution at all. The derivative of a cosine is a sine and thus I integrate the cosine function as if it were an x. Now, there is nothing wrong with substitution, I want to make that clear. But beyond Calc1, particularly in calc 2 and DFQ, we expect students to "see" the anti derivative if the chain rule is accounted for in an example worked out as I did above – imranfat Sep 23 '13 at 15:10