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Math Help - a very easy questin -integration

  1. #1
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    a very easy questin -integration

    how to do this piece?
    Attached Thumbnails Attached Thumbnails a very easy questin -integration-tan.gif  
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by afeasfaerw23231233 View Post
    how to do this piece?
    integrate by parts

    let u = \tan^2 x and v' = 1~dx

    and recall: \int uv' = uv - \int u'v
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  3. #3
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    The Pythagorean trig identity:
    sin^2(A) +cos^2(A) = 1
    Divide both sides by cos^2(A),
    tan^2(A) +1 = sec^2(A)
    tan^2(A) = sec^2(A) -1 --------------(i)

    So,
    INT.[tan^2(x)]dx
    = INT.[sec^2(x) -1]dx
    = tan(x) -x +C -------------------answer.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ticbol View Post
    The Pythagorean trig identity:
    sin^2(A) +cos^2(A) = 1
    Divide both sides by cos^2(A),
    tan^2(A) +1 = sec^2(A)
    tan^2(A) = sec^2(A) -1 --------------(i)

    So,
    INT.[tan^2(x)]dx
    = INT.[sec^2(x) -1]dx
    = tan(x) -x +C -------------------answer.
    very nice! and a lot easier than doing it by parts
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by afeasfaerw23231233 View Post
    how to do this piece?
    I would suggest:

    <br />
\int \tan^2(x)~dx=\int \frac{\sin^2(x)}{\cos^2(x)}~dx=\int \sin(x)~\frac{\sin(x)}{\cos^2(x)}~dx<br />

    Then try integration by parts

    RonL
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    thanks
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