- #1

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I've already subbed in:

[tex] x = 2tan\theta [/tex]

[tex] dx = 2sec^{2}\theta d\theta [/tex]

and I've gotten down to:

[tex] 16 \int tan^{3}\theta sec^{3}\theta d\theta [/tex]

But now I have noo idea what to do! Can someone give me a hint?

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- Thread starter Bazzinga
- Start date

- #1

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- 0

I've already subbed in:

[tex] x = 2tan\theta [/tex]

[tex] dx = 2sec^{2}\theta d\theta [/tex]

and I've gotten down to:

[tex] 16 \int tan^{3}\theta sec^{3}\theta d\theta [/tex]

But now I have noo idea what to do! Can someone give me a hint?

- #2

tiny-tim

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(where did your tan

sorry, but you just have to slog it out, using repeated integration by parts

substituting x = sinh(u) instead would have been a lot easier

- #3

rock.freak667

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[tex]

16 \int tan^{2}\theta sec^{3}\theta d\theta

[/tex]

as the x

Replace tan

- #4

SammyS

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...

and I've gotten down to:

[tex] 16 \int \tan^{3}\theta \sec^{3}\theta d\theta [/tex]

Hi Bazzinga.I think your integral should reduce to

[tex]

16 \int \tan^{2}\theta \sec^{3}\theta d\theta

[/tex]

as the x^{2}term outside the square root will give tan^{2}.

Replace tan^{2}θ with sec^{2}θ -1 and try to simplify it. I think the integrals may get a bit complicated.

The integral:

[tex] 16 \int \tan^{3}\theta \sec^{3}\theta \,d\theta [/tex]

works out very easily with

However, if

[tex]

16 \int \sec^{5}\theta -\sec^{3}\theta d\theta

[/tex]

- #5

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Oh sorry, the term is [tex]x^{3}[/tex] not [tex]x^{2}[/tex]

- #6

SammyS

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In that case,Oh sorry, the term is [tex]x^{3}[/tex] not [tex]x^{2}[/tex]

[tex]

\int \tan^{3}\theta \sec^{3}\theta \,d\theta = \int \tan\theta\,\left(\sec^{5}\theta-\sec^{3}\theta \right)\, d\theta

[/tex]

[tex]=\int \left(\sec^{4}\theta-\sec^{2}\theta \right)\tan\theta\,\sec\theta\ d\theta[/tex]

This is quite straight forward.

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