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#### Chipset3600

##### Member

- Feb 14, 2012

- 79

[TEX]\int tan(t)sec^3(t)dt[/TEX]

- Thread starter Chipset3600
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- #1

- Feb 14, 2012

- 79

[TEX]\int tan(t)sec^3(t)dt[/TEX]

- Mar 1, 2012

- 249

How do you mean "without using integration techniques"? Surely you need integration techniques to solve an integral? Also what have you tried?

[TEX]\int tan(t)sec^3(t)dt[/TEX]

Hint: Let $u = \sec(t) = \frac{1}{\cos(t)}$

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- #3

- Feb 14, 2012

- 79

I mean without: substitution, integration by parts...How do you mean "without using integration techniques"? Surely you need integration techniques to solve an integral? Also what have you tried?

Hint: Let $u = \sec(t) = \frac{1}{\cos(t)}$

- Feb 29, 2012

- 342

I'm not sure that is possible, this is not an elementary integral (in the sense that it is the derivative of an elementary function, such as $\cos x$, $\sin x$, $\tan x$ and so on).

Hello, Chipset3600!

How can i solve this without use integration technniques?

. . [TEX]\int \tan\theta \sec^3\!\theta\,d\theta[/TEX]

Well, maybe you can

If we have: .[tex]f(x) \:=\:\tfrac{1}{3}\sec^3\!\theta + C[/tex]

Then: .[tex]f'(x) \:=\:\tfrac{1}{3}\cdot 3\sec^2\!\theta\cdot\sec\theta\tan\theta + 0 \;=\;\tan\theta\sec^3\!\theta [/tex]

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- Feb 14, 2012

- 79

[TEX]\int tan(t).sec^3(t)dt = \int sec^2(t).sec(t).tan(t)dt[/TEX]Hello, Chipset3600!

Well, maybe you canall this?see

If we have: .[tex]f(x) \:=\:\tfrac{1}{3}\sec^3\!\theta + C[/tex]

Then: .[tex]f'(x) \:=\:\tfrac{1}{3}\cdot 3\sec^2\!\theta\cdot\sec\theta\tan\theta + 0 \;=\;\tan\theta\sec^3\!\theta [/tex]

Using the power rule now:

[TEX]\frac{sec^3(t)}{3}+C[/TEX]

- Feb 29, 2012

- 342

You don't have a polynomial to apply the power rule, at least not until you use substitution.