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Math Help - Quick integral question?

  1. #1
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    Question Quick integral question?

    Problem:
    Quick integral question?-15h041x.png

    I am stuck on this problem. Could it be as simple as changing the (tanx)^2 to (secx)^2
    and having the answer as (cosx)^2 - x?

    Or is it much more complicated than that? Please help me with this.
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  2. #2
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    \dfrac{1-tan^2(x)}{sec^2(x)}=cos^2(x)-sin^2(x)=cos(2x)
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  3. #3
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    Hello, florx!

    What you're saying is scary . . .


    \displaystyle \int\frac{1-\tan^2\!x}{\sec^2\!x}\,dx

    First of all: . \tan^2\!x \;=\;\sec^2\!x - 1

    \displaystyle\text{Substitute: }\;\frac{1 - (\sec^2\!x - 1)}{\sec^2\!x} \;=\;\frac{2-\sec^2\!x}{\sec^2\1x} \;=\; \frac{2}{\sec^2\!x} - \frac{\sec^2\!x}{\sec^2\!x}

    . . . . . . . =\;2\cos^2\!x - 1 \;=\;\cos2x .
    as Liverpool suggested
    Last edited by mr fantastic; January 28th 2011 at 04:43 AM. Reason: Leave something for the OP to do.
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