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    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!
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    (Original post by QueenOfNachos)
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    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.
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    (Original post by zetamcfc)
    Well, think of a substitution.
    Ah yes. How did I not think of that. Got it now, thanks a lot!!
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    (Original post by QueenOfNachos)
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    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?
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    (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!
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    (Original post by QueenOfNachos)
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    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" 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!
    Spoiler:
    Show
    \displaystyle \boxed{\therefore \int \sec^{2}\tan^{2}x \ \mathrm{d}x = \frac{\tan^{3}x}{3} + \mathrm{C}}
 
 
 
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