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Math Help - Integration

  1. #1
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    Integration

    please give the solution
    integrate sin3x/cos3x+sin3x from -pi by two to +
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by sareeta View Post
    please give the solution
    integrate sin3x/cos3x+sin3x from -pi by two to +
    \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{\sin(3x)}{\cos(3x)}+  \sin(3x)dx

    Rewrite this as \int\tan(3x)+\sin(3x)dx

    we must have the derivative of the quantity out side before we can do anythign

    so we have \frac{1}{3}\int{3\tan(3x)}dx+\frac{1}{3}\int{3\sin  (3x)}dx

    then by standard integration techniques we get

    \int\tan(3x)+\sin(3x)dx=\frac{-1}{3}\ln|\cos(3x)|-\frac{1}{3}\cos(3x)

    now just evaluate this from -pi/2 to pi/2
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  3. #3
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    Quote Originally Posted by sareeta View Post
    please give the solution
    integrate sin3x/cos3x+sin3x from -pi by two to +
    First I need to give thanks to the inte-killer for this trick

    \int\frac{\sin(3x)}{\cos(3x)+\sin(3x)}dx=\frac{1}{  2}\int\frac{2\sin(3x)}{\cos(3x)+\sin(3x)}dx=

    \frac{1}{2}\int\frac{2\sin(3x)+\cos(3x)-\cos(3x)}{\cos(3x)+\sin(3x)}dx=

    \frac{1}{2}\int\frac{\sin(3x)+\cos(3x)}{\sin(3x)+\  cos(3x)}dx+\frac{1}{2}\int \frac{\sin(3x)-\cos(3x)}{\sin(3x)+\cos(3x)}dx=

    \frac{1}{2}\int dx -\frac{1}{2}\int \frac{-\sin(3x)+\cos(3x)}{\sin(3x)+\cos(3x)}dx

    making a u sub on gives this as an antiderivative

    \frac{1}{2}x-\frac{1}{6}\ln|\sin(3x)+\cos(3x)|

    I'm not sure what you mean by your limits of integration.

    Good luck.
    Last edited by TheEmptySet; April 30th 2008 at 07:36 AM. Reason: Bad u sub thankss moo :)
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  4. #4
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    Quote Originally Posted by sareeta View Post
    please give the solution
    integrate sin3x/cos3x+sin3x from -pi by two to +
    There is a slicker solution since its a definite integral

    Let f(x) = \frac{\sin(x)}{\cos(x)+\sin(x)}

    So we want I = \int_{-\frac{\pi}2}^{\frac{\pi}2} f(3x) \, dx

    Lets u-sub  u = 3x and break the integral into two pieces,

    I_1 = \frac13 \int_{-\frac{3\pi}2}^{0} f(u) \, du and I_2 = \frac13 \int_{0}^{\frac{3\pi}2} f(u) \, du


    I_1 = \frac13 \int_{-\frac{3\pi}2}^{0} f(u) \, du = \frac13 \int_{-\frac{3\pi}2}^{0} f\left(-\frac{3\pi}2 - u\right) \, du = <br />
\frac13 \int_{-\frac{3\pi}2}^{0} (1 - f(u)) \, du

    \Rightarrow 2I_1 = \frac13 \int_{-\frac{3\pi}2}^{0}  \, du = \frac13 .\frac{3\pi}2<br />
\Rightarrow  I_1 = \frac{\pi}4

    Clearly I_2 - I_1 = 0 \Rightarrow I_1 = I_2 = \frac{\pi}4
    So I = I_1 + I_2 = \frac{\pi}2
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  5. #5
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    Hello,

    Quote Originally Posted by TheEmptySet View Post
    First I need to give thanks to the inte-killer for this trick

    \int\frac{\sin(3x)}{\cos(3x)+\sin(3x)}dx=\frac{1}{  2}\int\frac{2\sin(3x)}{\cos(3x)+\sin(3x)}dx=

    \frac{1}{2}\int\frac{2\sin(3x)+\cos(3x)-\cos(3x)}{\cos(3x)+\sin(3x)}dx=

    \frac{1}{2}\int\frac{\sin(3x)+\cos(3x)}{\sin(3x)+\  cos(3x)}dx+\frac{1}{2}\int \frac{\sin(3x)-\cos(3x)}{\sin(3x)+\cos(3x)}dx=

    \frac{1}{2}\int dx -\frac{1}{2}\int \frac{-\sin(3x)+\cos(3x)}{\sin(3x)+\cos(3x)}dx

    making a u sub on gives this as an antiderivative

    \frac{1}{2}x-\frac{1}{\color{red}6}\ln|\sin(3x)+\cos(3x)|

    I'm not sure what you mean by your limits of integration.

    Good luck.
    There's a 3 missing (red)

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