1. ## 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

2. $\displaystyle \int\limits_{ - 2}^{ - 1} {\left[ {\frac{1} {{\sin \left( {2\theta } \right)}} - \cot \left( {2\theta } \right)} \right]} ^2 d\theta$

$\displaystyle = \int\limits_{ - 2}^{ - 1} {\left[ {\frac{{1 - \cos \left( {2\theta } \right)}} {{\sin \left( {2\theta } \right)}}} \right]} ^2 d\theta = \int\limits_{ - 2}^{ - 1} {\left[ {\frac{{2\sin ^2 \theta }} {{\sin \left( {2\theta } \right)}}} \right]} ^2 d\theta = \int\limits_{ - 2}^{ - 1} {\left[ {\frac{{\sin \theta }} {{\cos \theta }}} \right]} ^2 d\theta$

$\displaystyle = \int\limits_{ - 2}^{ - 1} {\frac{{1 - \cos ^2 \theta }} {{\cos ^2 \theta }}} d\theta = \left. {\tan \theta - \theta } \right|_{ - 2}^{ - 1}$
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$\displaystyle \begin{gathered} \int {\frac{1} {{\sin x}}} dx = \int {\frac{{\sin x}} {{\sin ^2 x}}} dx \hfill \\ \hfill \\ t = \cos x \hfill \\ dt = - \sin xdx \hfill \\ \end{gathered}$

$\displaystyle = - \int {\frac{1} {{1 - t^2 }}} dx$

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3. Hello, jarny!

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

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

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

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

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

I let you evaluate it . . .

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

show that: .$\displaystyle -\ln|\csc x+\cot x| \:= \: \ln|\csc x-\cot x|$

We have: .$\displaystyle -\ln(\csc x + \cot x) \;=\;\ln(\csc x + \cot x)^{-1} \;=\;\ln\left(\frac{1}{\csc x + \cot x}\right)$

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

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

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

4. Thanks alot guys. I appreciate it.