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What is Integral tan 2x dx?

  1. Feb 10, 2016 #1
    What is ##\int \tan 2x \ dx##?

    What I get is

    ##\int \tan 2x \ dx##
    ##= \int \frac{\sin 2x}{\cos 2x} dx##
    ##= \int \frac{2 \sin x \cos x}{1 - 2 \sin^2 x}dx##

    let u = sin x then ##\frac{du}{dx} = \cos x## or du = cos x dx

    So

    ##= \int \frac{2 \sin x \cos x}{1 - 2 \sin^2 x}dx##
    ##= \int \frac{2u}{1 - 2u^2} du##
     
  2. jcsd
  3. Feb 10, 2016 #2

    Samy_A

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    There may be easier ways to do this, but since you tried it this way, notice that ##2u du= -\frac{1}{2}d(1-2u²)##
     
  4. Feb 10, 2016 #3

    SteamKing

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    Rather than saying u = sin x, use u = 2x instead. Just expand tan u into ##\frac{sin\, u}{cos\, u}##.
    This integral is much easier to solve.

    Expanding sin 2x and cos 2x in terms of sin x and cos x just makes things more complicated.
     
  5. Feb 11, 2016 #4

    HallsofIvy

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    Let ##v= 1- 2u^2## so that ##dv= -4u du##. Then ##(-1/2)dv= 2u du## and the integral becomes ##-\frac{1}{2}\int \frac{dv}{v}##
     
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