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Math Help - integration

  1. #1
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    integration

    how do we integrate

    (sin2 x)/cos x
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  2. #2
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: integration

    Quote Originally Posted by priya190992 View Post
    how do we integrate

    (sin2 x)/cos x
    Try: \int \frac{\text{Sin}(x)^2}{\text{Cos}(x)} \, dx=\int -\frac{1}{2} (-1+\text{Cos}(2 x)) \text{Sec}(x)dx
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  3. #3
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    Re: integration

    \int \frac{\sin^2(x)}{\cos(x)} dx

    = \int \frac{\sin^2(x)}{\cos^2(x)} \cos(x) dx

    = \int \frac{\sin^2(x)}{1 - \sin^2(x)} \cos(x) dx

    = \int \frac{u^2}{1 - u^2} du (where u = \sin(x)).

    = \int \frac{u^2 - 1+1}{1 - u^2} du

    = \int \left( \frac{u^2 - 1}{1 - u^2} + \frac{1}{1 - u^2} \right) du

    = \int (-1 + \frac{1}{1 - u^2}) du

    = -\int du + \int \frac{1}{1 - u^2} du

    = -u + \int \frac{1}{(1 - u)(1+u)} du

    = -u + \int \left( \frac{1/2}{(1 - u)} + \frac{1/2}{(1+u)} \right) du

    = -u + (1/2) \int \frac{du}{(1 - u)} + (1/2) \int \frac{du}{(1+u)}

    = -u + (1/2) (-\ln|1-u|) + (1/2) (\ln|1+u|) + C

    = -u + (1/2) \ln\left( \left\lvert \frac{1+u}{1-u} \right\rvert \right) + C

    = -\sin(x) + (1/2) \ln\left( \left\lvert \frac{1+\sin(x)}{1-\sin(x)} \right\rvert \right) + C

    = -\sin(x) + (1/2) \ln\left( \left\lvert \frac{(1+\sin(x))^2}{\cos^2(x)} \right\rvert \right) + C

    = -\sin(x) + (1/2) \ln\left( \left\lvert \frac{1+\sin(x)}{\cos(x)} \right\rvert^2 \right) + C

    = -\sin(x) + \ln\left( \left\lvert \frac{1+\sin(x)}{\cos(x)} \right\rvert \right) + C.

    = -\sin(x) + \ln\left( \frac{1+\sin(x)}{|\cos(x)|} \right) + C.

    *** Check ***

    On a domain where \cos(x) > 0:

    \frac{d}{dx}\left\{ -\sin(x) + \ln\left( \frac{(1+\sin(x))}{\cos(x)} \right) + C \right\}

    = -(\cos(x)) + \frac{\cos(x)}{(1+\sin(x))} \frac{d}{dx} \left\{ \frac{(1+\sin(x))}{\cos(x)} \right\} + (0)

    = -\cos(x) + \frac{\cos(x)}{(1+\sin(x))} \left( \frac{(\cos(x))(\cos(x))-(1+\sin(x))(-\sin(x))}{\cos^2(x)} \right)

    = -\cos(x) + \frac{\cos(x)}{(1+\sin(x))} \left( \frac{\cos^2(x)+\sin^2(x) + \sin(x)}{\cos^2(x)} \right)

    = -\cos(x) + \frac{\cos(x)}{(1+\sin(x))} \left( \frac{1 + \sin(x)}{\cos^2(x)} \right)

    = -\cos(x) + \frac{1}{\cos(x)} = -\frac{\cos^2(x)}{\cos(x)} + \frac{1}{\cos(x)} = \frac{1 - \cos^2(x)}{\cos(x)}

    = \frac{\sin^2(x)}{\cos(x)}.
    Last edited by johnsomeone; September 21st 2012 at 05:23 PM.
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  4. #4
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    Re: integration

    Hello, priya190992!

    \displaystyle \int\frac{\sin^2\!x}{\cos x}\,dx

    We have: . \displaystyle\int\frac{1-\cos^2\!x}{\cos x}\,dx \;=\;\int\left(\frac{1}{\cos x} - \cos x\right)dx \;=\;\int(\sec x - \cos x)\,dx

    . . . =\;\ln|\sec x + \tan x| - \sin x + C
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