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Math Help - trig integral

  1. #1
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    trig integral

    \int{sec^4xtan^4x}

    I know the answer is \frac{1}{9}tan^9x+\frac{1}{7}tan^7x +C

    But what if I had
    \int{{sec^4}3x{tan^4}3x}

    Would the answer just be

    \frac{1}{9}{tan^9}3x+\frac{1}{7}{tan^7}3x +C
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by gammaman View Post
    \int{sec^4xtan^4x}

    I know the answer is \frac{1}{9}tan^9x+\frac{1}{7}tan^7x +C

    But what if I had
    \int{{sec^4}3x{tan^4}3x}

    Would the answer just be

    \frac{1}{9}{tan^9}3x+\frac{1}{7}{tan^7}3x +C
    No, it would be 1/3 of that answer plus the constant term, because you are using u-substitution with u = 3x, so du = 3dx inside the integral, hence you must divide by 3 in front of the integral sign. good luck!!
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  3. #3
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    Hello, gammaman!

    Good thinking . . . but it's slightly wrong.


    \int \sec^4\!x\tan^4\!x\,dx \quad\Rightarrow\quad \tfrac{1}{9}\tan^9\!x+\tfrac{1}{7}\tan^7\!x +C

    But what if I had: . \int \sec^4\!3x\tan^4\!3x\,dx

    Would the answer just be: . \tfrac{1}{9}\tan^9\!3x+\tfrac{1}{7}\tan^7\!3x +C . . . . no

    Because of the 3x, we'd use: . u \:=\:3x\quad\Rightarrow\quad du \:=\:3\,dx \quad\Rightarrow\quad dx \:=\:\tfrac{1}{3}\,du

    Substitute: . \int\sec^4\!u\tan^4\!u\left(\tfrac{1}{3}\,du\right  ) \;=\;\tfrac{1}{3}\int\sec^4\!u\tan^4\!u\,du


    We get the same answer ... with \tfrac{1}{3} in front.

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