Paul's Online Notes
Home / Calculus II / Integration Techniques / Integrals Involving Trig Functions
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 7.2 : Integrals Involving Trig Functions

11. Evaluate $$\displaystyle \int{{{{\csc }^6}\left( {\frac{1}{4}w} \right){{\cot }^4}\left( {\frac{1}{4}w} \right)\,dw}}$$.

Show All Steps Hide All Steps

Hint : Even though no examples of products of cotangents and cosecants were done in the notes for this section you should know how to do them. Ask yourself how you would do the problem if it involved tangents and secants instead and you should be able to see how to do this problem as well.
Start Solution

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 cosecant is even and so we can strip out two of them.

$\int{{{{\csc }^6}\left( {\frac{1}{4}w} \right){{\cot }^4}\left( {\frac{1}{4}w} \right)\,dw}} = \int{{{{\csc }^4}\left( {\frac{1}{4}w} \right){{\cot }^4}\left( {\frac{1}{4}w} \right)\,\,\,{{\csc }^2}\left( {\frac{1}{4}w} \right)\,dw}}$ Show Step 2

Now we can use the trig identity $${\cot ^2}\theta + 1 = {\csc ^2}\theta$$ to convert the remaining cosecants to cotangents.

\begin{align*}\int{{{{\csc }^6}\left( {\frac{1}{4}w} \right){{\cot }^4}\left( {\frac{1}{4}w} \right)\,dw}} & = \int{{{{\left[ {{{\csc }^2}\left( {\frac{1}{4}w} \right)} \right]}^2}{{\cot }^4}\left( {\frac{1}{4}w} \right)\,\,\,{{\csc }^2}\left( {\frac{1}{4}w} \right)\,dw}}\\ & = \int{{{{\left[ {{{\cot }^2}\left( {\frac{1}{4}w} \right) + 1} \right]}^2}{{\cot }^4}\left( {\frac{1}{4}w} \right)\,\,\,{{\csc }^2}\left( {\frac{1}{4}w} \right)\,dw}}\end{align*} Show Step 3

Now we can use the substitution $$u = \cot \left( {\frac{1}{4}w} \right)$$ to evaluate the integral.

\begin{align*}\int{{{{\csc }^6}\left( {\frac{1}{4}w} \right){{\cot }^4}\left( {\frac{1}{4}w} \right)\,dw}} & = - 4\int{{{{\left[ {{u^2} + 1} \right]}^2}{u^4}\,du}}\\ & = - 4\int{{{u^8} + 2{u^6} + {u^4}\,du}} = - 4\left( {\frac{1}{9}{u^9} + \frac{2}{7}{u^7} + \frac{1}{5}{u^5}} \right) + c\end{align*}

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 4

Don’t forget to substitute back in for $$u$$!

$\int{{{{\csc }^6}\left( {\frac{1}{4}w} \right){{\cot }^4}\left( {\frac{1}{4}w} \right)\,dw}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{4}{9}{{\cot }^9}\left( {\frac{1}{4}w} \right) - \frac{8}{7}{{\cot }^7}\left( {\frac{1}{4}w} \right) - \frac{4}{5}{{\cot }^5}\left( {\frac{1}{4}w} \right) + c}}$