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Math Help - Integration

  1. #1
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    Integration

    i am having trouble with two problems. 1) S(lower limit -2 upper limit -1) of (csc(2theta)-cot(2theta))^2 dtheta and 2) From Scscx dx=
    -ln|cscx+cotx|+c Scscx dx= ln|cscx-cotx|+c show -ln|cscx+cotx|=
    ln|cscx-cotx|+c ....................i don't know what it is about the cscx/cotx combo im having trouble with but id appreciate some help. Thanks
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  2. #2
    Senior Member Peritus's Avatar
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    <br />
\int\limits_{ - 2}^{ - 1} {\left[ {\frac{1}<br />
{{\sin \left( {2\theta } \right)}} - \cot \left( {2\theta } \right)} \right]} ^2 d\theta


     = \int\limits_{ - 2}^{ - 1} {\left[ {\frac{{1 - \cos \left( {2\theta } \right)}}<br />
{{\sin \left( {2\theta } \right)}}} \right]} ^2 d\theta  = \int\limits_{ - 2}^{ - 1} {\left[ {\frac{{2\sin ^2 \theta }}<br />
{{\sin \left( {2\theta } \right)}}} \right]} ^2 d\theta  = \int\limits_{ - 2}^{ - 1} {\left[ {\frac{{\sin \theta }}<br />
{{\cos \theta }}} \right]} ^2 d\theta

    <br /> <br />
 = \int\limits_{ - 2}^{ - 1} {\frac{{1 - \cos ^2 \theta }}<br />
{{\cos ^2 \theta }}} d\theta  = \left. {\tan \theta  - \theta } \right|_{ - 2}^{ - 1} <br /> <br />
    ------------------------------------------------------------------------

    <br />
\begin{gathered}<br />
  \int {\frac{1}<br />
{{\sin x}}} dx = \int {\frac{{\sin x}}<br />
{{\sin ^2 x}}} dx \hfill \\<br />
   \hfill \\<br />
  t = \cos x \hfill \\<br />
  dt =  - \sin xdx \hfill \\ <br />
\end{gathered} <br />

    <br />
 =  - \int {\frac{1}<br />
{{1 - t^2 }}} dx<br />

    .....
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  3. #3
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    Hello, jarny!

    1)\;\;\int^{-1}_{-2}(\csc2\theta-\cot2\theta)^2\,d\theta

    The integrand is: . (\csc2\theta - \cot2\theta)^2 \:=\:\csc^2\!2\theta - 2\csc\theta\cot\theta + \cot^2\!2\theta

    . .  = \;\csc^2\!2\theta - 2\csc\theta\cot\theta + (\csc^2\!2\theta - 1) \;=\;2\csc^2\!2\theta - 2\csc2\theta\cot2\theta - 1


    \text{The integral is: }\;\underbrace{\int2\csc^2\!2\theta\,d\theta} - \underbrace{\int2\csc2\theta\cot2\theta\,d\theta} - \underbrace{\int d\theta}

    . . . . . . . . = \quad-\cot2\theta \qquad+ \qquad\csc2\theta \qquad- \qquad\theta \quad+ \quad C


    I let you evaluate it . . .



    2) From:. \int\csc x\,dx \;=\;\begin{array}{c}\text{-}\ln|\csc x+\cot x|+C \\ \ln|\csc x-\cot x|+C \end{array}

    show that: . -\ln|\csc x+\cot x| \:= \:<br />
\ln|\csc x-\cot x|

    We have: . -\ln(\csc x + \cot x) \;=\;\ln(\csc x + \cot x)^{-1} \;=\;\ln\left(\frac{1}{\csc x + \cot x}\right)<br />


    Multiply by \frac{\csc x - \cot x}{\csc x - \cot x}\!:\;\;\ln\left(\frac{1}{\csc x + \cot x}\cdot\frac{csc x - \cot x}{\csc x - \cot x}\right) \;=\;\ln\left(\frac{\csc x - \cot x}{\csc^2\!x-\cot^2\!x}\right)


    Recall the identity: . \csc^2\!\theta - \cot^2\!\theta \:=\:1

    Hence, we have: . \ln\left(\frac{\csc x - \cot x}{1}\right) \;=\;\ln(\csc x - \cot x)\quad\hdots\quad \text{ta-}DAA!

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  4. #4
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    Thanks alot guys. I appreciate it.
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