Section 7.2 : Integrals Involving Trig Functions
10. Evaluate \( \displaystyle \int{{\cot \left( {10z} \right){{\csc }^4}\left( {10z} \right)\,dz}}\).
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Other than the obvious difference in the actual functions there is no practical difference in how this problem and one that had tangents and secants would work. So, all we need to do is ask ourselves how this would work if it involved tangents and secants and we’ll be able to work this on as well.
We can first notice here is that the exponent on the cotangent is odd and we’ve got a cosecant in the problems and so we can strip the (only) cotangent and one of the secants out.
\[ \int{{\cot \left( {10z} \right){{\csc }^4}\left( {10z} \right)\,dz}} = \int{{{{\csc }^3}\left( {10z} \right)\,\,\,\,\cot \left( {10z} \right)\csc \left( {10z} \right)\,dz}}\] Show Step 2Normally we would use the trig identity \({\cot ^2}\theta + 1 = {\csc ^2}\theta \) to convert the remaining cotangents to cosecants. However, in this case there are no remaining cotangents to convert and so there really isn’t anything to do at this point other than to use the substitution \(u = \csc \left( {10z} \right)\) to evaluate the integral.
\[ \int{{\cot \left( {10z} \right){{\csc }^4}\left( {10z} \right)\,dz}} = - \frac{1}{{10}}\int{{{u^3}\,du}} = - \frac{1}{{40}}{u^4} + c\]Note that we’ll not be doing the actual substitution work here. At this point it is assumed that you recall substitution well enough to fill in the details if you need to. If you are rusty on substitutions you should probably go back to the Calculus I practice problems and practice on the substitutions.
Show Step 3Don’t forget to substitute back in for \(u\)!
\[ \int{{\cot \left( {10z} \right){{\csc }^4}\left( {10z} \right)\,dz}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{1}{{40}}{{\csc }^4}\left( {10z} \right) + c}}\]