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Math Help - integration

  1. #1
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    integration

    How would you take the integral of sec^2(3x)?
    The ^2 thing messes me up.
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  2. #2
    MHF Contributor red_dog's Avatar
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    \displaystyle \int\sec ^23xdx=\int\frac{dx}{\cos ^23x}=\frac{1}{3}\int \frac{3}{\cos ^23x}dx=\frac{1}{3}\tan 3x+C.

    Or by substitution 3x=t\Rightarrow 3dx=dt\Rightarrow dx=\frac{1}{3}dt
    Then \frac{1}{3}\int\frac{dt}{\cos ^2t}=\frac{1}{3}\tan t+C.
    Back substitution \frac{1}{3}\tan 3x+C
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  3. #3
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    Quote Originally Posted by red_dog View Post
    \displaystyle \int\sec ^23xdx=\int\frac{dx}{\cos ^23x}=\frac{1}{3}\int \frac{3}{\cos ^23x}dx=\frac{1}{3}\tan 3x+C.

    Or by substitution 3x=t\Rightarrow 3dx=dt\Rightarrow dx=\frac{1}{3}dt
    Then \frac{1}{3}\int\frac{dt}{\cos ^2t}=\frac{1}{3}\tan t+C.
    Back substitution \frac{1}{3}\tan 3x+C
    how did you get from 3/cos^2(3x) = tan3x?
    tan is sin/cos, but there isnt a sin
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  4. #4
    MHF Contributor red_dog's Avatar
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    (\tan x)'=\frac{1}{\cos ^2x}\Rightarrow \int\frac{1}{\cos ^2x}dx=\tan x+C
    (\tan u)'=\frac{u'}{\cos ^2u}\Rightarrow \int\frac{u'}{\cos ^2u}du=\tan u+C
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