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C4 Integration

I am not sure why the integral of sec^(2)xtan^(2)x is 1/3tan^(3)x.
Please can anyone show me the steps to get there. Much appreciated!

Reply 1

zetamcfc

Original post by QueenOfNachos

I am not sure why the integral of sec^(2)xtan^(2)x is 1/3tan^(3)x.
Please can anyone show me the steps to get there. Much appreciated!

Well, think of a substitution.

Reply 2

QueenOfNachos

Original post by zetamcfc

Well, think of a substitution.

Ah yes. How did I not think of that. Got it now, thanks a lot!! :biggrin:

Reply 3

atsruser

Original post by QueenOfNachos

I am not sure why the integral of sec^(2)xtan^(2)x is 1/3tan^(3)x.
Please can anyone show me the steps to get there. Much appreciated!

If you start with

y=\tan^3 x

and write

u=\tan x \Rightarrow y=u^3

then by the chain rule:

\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = 3u^2 \times \sec^2 x

.

Can you finish this?

Reply 4

QueenOfNachos

Original post by atsruser

If you start with

y=\tan^3 x

and write

u=\tan x \Rightarrow y=u^3

then by the chain rule:

\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = 3u^2 \times \sec^2 x

.

Can you finish this?

Yes I can, thank you very much! :smile:

Reply 5

Ayman!

Original post by QueenOfNachos

I am not sure why the integral of sec^(2)xtan^(2)x is 1/3tan^(3)x.
Please can anyone show me the steps to get there. Much appreciated!

I'm a bit late to the party, but I prefer to do these integrals by recognition (also referred to as "reverse chain rule" :wink:

rather than substitution - it's a little quicker than using a sub. I'll show you what I mean.

We know that

\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\left ( \tan^{3}x \right ) = 3\sec^{2}x\tan^{2}x

via the chain rule, right? Since we know the derivative of this, we can say that

\displaystyle \Rightarrow \int 3 \sec^{2}x\tan^{2}x \ \mathrm{d}x = \tan^{3}x + \mathrm{C_{1}}

Won't be too tough finding the integral from here, I hope! :smile:

Spoiler

(edited 8 years ago)

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C4 Integration

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