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How Can I evaluate $\displaystyle \int \sec^3 (x)dx$

(Without Using Weierstrass Substution or Integration by parts.)

$\bf{My\; Try::}$ Let $\displaystyle I = \int\sec^3(x)dx = \int \frac{1}{\cos^3(x)}dx = \int \frac{1}{\sin ^3\left(\frac{\pi}{2}-x\right)}dx$

Now Let $\displaystyle \left(\frac{\pi}{2}-x\right) = t\;,$ Then $\displaystyle dx = -dt$. So $\displaystyle I = -\int \frac{1}{\sin^3 t}dt = -\int\frac{1}{2\sin^3\left(\frac{t}{2}\right)\cdot \cos^3 \left(\frac{t}{2}\right)}dt$

Now How can I solve after that

Help me

Thanks

juantheron
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2 Answers2

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If you don't mind partial fractions, you can simply rewrite your integral as

$$\int\frac{\cos x}{\cos^4x}dx=\int\frac{\cos x}{(1-\sin^2x)^2}dx$$

then substitute $u=\sin x$.

Mike
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    Anyone care to explain what's with all the downvotes? – Mike Aug 19 '14 at 19:33
  • Your method looks completely fine to me, though it is not the standard way of going about it. I would say that it is actually fairly clever in the way that it makes use of trigonometric identities. – Gahawar Aug 19 '14 at 19:40
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    @Mike I haven't a clue, but it's a very clever way of evaluating the given integral, rather than the standard Weierstrass substitution. – beep-boop Aug 19 '14 at 20:07
  • @Gahawar I can't take credit for the method. I remember something similar from a previous topic. It appears this whole topic got hit with a flood of downvoters. – Mike Aug 19 '14 at 22:36
  • @Gahawar: To me, this is the standard way of integrating odd powers of sin or cos... – Hans Lundmark Aug 20 '14 at 00:31
  • @HansLundmark Certainly, I meant that one does not typically use such a method when integrating $\sec^3 x$. At least in my experience, I have seen the method used by user156227 more frequently as it requires slightly less work, though both do the job equally well. – Gahawar Aug 20 '14 at 01:06
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Hint:

Substitite

$$\sec (x)= \cosh (u) $$

$$\tan(x) = \sinh(u)$$

$$\sec^2 (x) \, \mathrm dx = \cosh (u) \,\mathrm du$$

$$\mathrm dx = \frac1{\cosh(u)} \, \mathrm du$$

Edit: Why was this downvoted?

Here is the entire process

$$\begin{align} \int \sec^3 x \, \mathrm dx &{}= \int \cosh^2 u\,\mathrm du \\ &= \frac{1}{2}\int ( \cosh 2u +1) \,\mathrm du \\ &= \frac{1}{2} \left( \frac{1}{2}\sinh2u + u\right) + C\\ &= \frac{1}{2} ( \sinh u \cosh u + u ) + C \\ &= \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln|\sec x + \tan x| + C \end{align} $$

user157227
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  • Your answer is correct. Upvote given! As for the downvote, it may not be obvious from inspection that $dx = \sech(u) du$. I would advise writing that it is true since $d[\sec(x)] = \tan(x)\cdot \sec(x) dx = \sinh(u)\cdot \cosh(u); dx$, whence... Or something along those lines. – Christopher K Aug 19 '14 at 22:00
  • @Chris Thanks for the advice, I have now shown some more steps. – user157227 Aug 19 '14 at 22:08