Question 1121444
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int^{\small{\pi/4}}_{\small{\pi/6}}\LARGE\,\csc^3(x)\cot(x)dx]


Let *[tex \Large u\ =\ \csc^3(x)], then *[tex \Large \frac{du}{dx}\ =\ 3\csc^2(x)\frac{d}{dx}\csc(x)\ =\ -3\cot(x)\csc(x)\csc^2(x)\ =\ -3\cot(x)\csc^3(x)].  Therefore *[tex \Large dx\ =\ \frac{1}{-3\cot(x)\csc^3(x)}du].  Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int\,\csc^3(x)\cot(x)dx\ =\ -\frac{1}{3}\int\,du\ =\ -\frac{1}{3}u\ +\ C]


Reversing the substitution:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\frac{1}{3}u\ =\ -\frac{\csc^3(x)}{3}]


The rest is just arithmetic.  Hint: *[tex \Large \csc] is the reciprocal of *[tex \Large \sin] and you should know the values for *[tex \Large \sin(\pi/4)] and *[tex \Large \sin(\pi/6)] by now. 
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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