Section 5.7 : Computing Definite Integrals
12. Evaluate the following integral, if possible. If it is not possible clearly explain why it is not possible to evaluate the integral.
\[\int_{{\frac{\pi }{6}}}^{{\frac{\pi }{3}}}{{2{{\sec }^2}\left( w \right) - 8\csc \left( w \right)\cot \left( w \right)\,dw}}\]Show All Steps Hide All Steps
Start SolutionFirst notice that even though we do have some “hidden” rational expression here (in the definitions of the trig functions) neither cosine nor sine is zero in the interval we are integrating over and so both terms are continuous over the interval.
Therefore all we need to do integrate the function.
\[\int_{{\frac{\pi }{6}}}^{{\frac{\pi }{3}}}{{2{{\sec }^2}\left( w \right) - 8\csc \left( w \right)\cot \left( w \right)\,dw}} = \left. {\left( {2\tan \left( w \right) + 8\csc \left( w \right)} \right)} \right|_{\frac{\pi }{6}}^{\frac{\pi }{3}}\]Recall that we don’t need to add the “+c” in the definite integral case as it will just cancel in the next step.
Show Step 2The final step is then just to do the evaluation.
We’ll leave the basic arithmetic to you to verify and only show the results of the evaluation. Make sure that you evaluate the upper limit first and then subtract off the evaluation at the lower limit.
Here is the answer for this problem.
\[\int_{{\frac{\pi }{6}}}^{{\frac{\pi }{3}}}{{2{{\sec }^2}\left( w \right) - 8\csc \left( w \right)\cot \left( w \right)\,dw}} = \left( {\frac{{16}}{{\sqrt 3 }} + 2\sqrt 3 } \right) - \left( {16 + \frac{2}{{\sqrt 3 }}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{14}}{{\sqrt 3 }} + 2\sqrt 3 - 16}}\]