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Math Help - Integration with trig functions

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    Integration with trig functions

    Evaluate the integral sinx/cos^2x dx from 0 to pi/3.


    So I've gotten it to tanxcosx, but I don't know the indefinite integral of these, and I couldn't find it in my book. Can anyone help?
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    Senior Member yeKciM's Avatar
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    Quote Originally Posted by bobsanchez View Post
    Evaluate the integral sinx/cos^2x dx from 0 to pi/3.


    So I've gotten it to tanxcosx, but I don't know the indefinite integral of these, and I couldn't find it in my book. Can anyone help?
    you should get under integral  \sec{x} \tan{x} \; dx .....
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    \displaystyle \int {\frac{{\sin (x)dx}}<br />
{{\cos ^2 (x)}} = \frac{{ 1}}<br />
{{\cos (x)}}}
    Last edited by Plato; August 11th 2010 at 01:37 PM.
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    Ah okay, let me work the problem. Thanks.
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    -1?
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    Senior Member yeKciM's Avatar
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    i got result 1
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    -1/cos(60) = -2

    -1/cos(0) = -1

    -2 - (-1) = -2 + 1 = -1

    Did I mess up somewhere in there or enter it wrong?
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    Let u = \cos{x} to get Plato's result (without the typo minus).
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    Senior Member yeKciM's Avatar
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    Quote Originally Posted by bobsanchez View Post
    -1/cos(60) = -2

    -1/cos(0) = -1

    -2 - (-1) = -2 + 1 = -1

    Did I mess up somewhere in there or enter it wrong?

     \displaystyle \int _0 ^{\frac {\pi}{3} } \frac {\sin{x}}{\cos^2{x}} \; dx =  \int _0 ^{\frac {\pi}{3} } \sec{x} \tan{x} \; dx =

     \displaystyle u = \sec {x}  and now the  \displaystyle du= \tan{x}\sec{x}\;  dx

    after that ... i just got

     \displaystyle \int _0 ^{\frac {\pi}{3} }\;du = u \big{|} _0 ^{\frac {\pi}{3} }= \sec{x}  \big{|} _0 ^{\frac {\pi}{3} }


    and with limits you get "1"
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    Evaluate the integral sec2xtan2x dx.
    sec2x + C


    This is another good one. My first try resulted in sec2x + C, but was wrong.
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    Quote Originally Posted by bobsanchez View Post
    Evaluate the integral sinx/cos^2x dx from 0 to pi/3.


    So I've gotten it to tanxcosx, but I don't know the indefinite integral of these, and I couldn't find it in my book. Can anyone help?
    Just use a u substitution...

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

    Let u=\cos{x} so that du = -\sin{x}\,dx and the integral becomes

    -\int{u^{-2}\,du}

     = -\left(\frac{u^{-1}}{-1}\right) + C

     = \frac{1}{u} + C

     = \frac{1}{\cos{x}} + C.


    Therefore \int_0^{\frac{\pi}{3}}{\frac{\sin{x}}{\cos^2{x}}\,  dx} = \left[\frac{1}{\cos{x}}\right]_0^{\frac{\pi}{3}}

     = \left[\frac{1}{\cos{\left(\frac{\pi}{3}\right)}}\right] - \left[\frac{1}{\cos{(0)}}\right]

     = \frac{1}{\frac{1}{2}} - \frac{1}{1}

     = 2 - 1

     = 1.
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    Quote Originally Posted by bobsanchez View Post
    Evaluate the integral sec2xtan2x dx.
    sec2x + C


    This is another good one. My first try resulted in sec2x + C, but was wrong.
    Can someone take a quick look at this one please?
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    Senior Member eumyang's Avatar
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    Quote Originally Posted by bobsanchez View Post
    Can someone take a quick look at this one please?
    You forgot to use the chain rule when finding du.

    \begin{aligned}<br />
u &= \sec 2x \\<br />
du &= 2\sec 2x\, \tan 2x \, dx \\<br />
\frac{1}{2}du &= \sec 2x\, \tan 2x \, dx<br />
\end{aligned}


    \displaystyle \int \sec 2x\,\tan 2x\, dx
    = \displaystyle \int \frac{1}{2} du
    Can you take it from here?
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    ...wow...forgot the chain rule? Really? *sigh*

    Anyways, thanks for the help. Alright, so...I just integrate and multiply it by du, correct? So (1/2)sec2x + C?
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    Senior Member eumyang's Avatar
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    I could read that as \sec \frac{2x}{2} + C, but I know that's not what you meant.

     \displaystyle \int \frac{1}{2} du = \frac{1}{2}u + C = \frac{1}{2}\sec 2x + C
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