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Thread: [SOLVED] Verify trig substitution

  1. #1
    Senior Member mollymcf2009's Avatar
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    [SOLVED] Verify trig substitution

    Hey y'all!

    Here is my integral:

    $\displaystyle 75 \int sec^2x tanx dx$

    The book does the substitution with $\displaystyle u = tanx$, $\displaystyle du = sec^2x dx$

    I worked mine out before I looked at the answer, I changed everything to terms of sin and cos and worked it that way.

    My answer:

    $\displaystyle \frac{75}{2cos^2x} + C$

    Book answer:

    $\displaystyle \frac{75}{2} tan^2x + C$

    Is my answer wrong?
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  2. #2
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    Nope you're perfectly fine.

    To see why, use the fact that: $\displaystyle \sec^2 x = \tan^2 x + 1$

    $\displaystyle \begin{aligned} \frac{75}{2\cos^2 x} + C & = \frac{75}{2}\sec^2 x + C \\ & = \frac{75}{2} \left(\tan^2 x + 1\right) + C \\ & = \frac{75}{2} \tan^2 x + {\color{red}\frac{75}{2} + C} \end{aligned}$

    The stuff in red is nothing more than a constant, so call it $\displaystyle K$

    So your expression is equal to: $\displaystyle \frac{75}{2}\tan^2 x + K$

    which is exactly what the book had.

    Remember, all antiderivatives of a given function differ only by their constants which is what we've shown here.
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