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Section 7.2 : Integrals Involving Trig Functions

14. Evaluate \( \displaystyle \int{{\left[ {9{{\sin }^5}\left( {3x} \right) - 2{{\cos }^3}\left( {3x} \right)} \right]{{\csc }^4}\left( {3x} \right)\,dx}}\).

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Hint : Since this has a mix of trig functions maybe the best option would be to first get it reduced down to just a couple that we know how to deal with.
Start Solution

To get started on this problem we should first probably see if we can reduce the integrand down to just sines and cosines. This is easy enough to do simply by recalling the definition of cosecant in terms of sine.

\[\begin{align*}\int{{\left[ {9{{\sin }^5}\left( {3x} \right) - 2{{\cos }^3}\left( {3x} \right)} \right]{{\csc }^4}\left( {3x} \right)\,dx}} & = \int{{\left[ {9{{\sin }^5}\left( {3x} \right) - 2{{\cos }^3}\left( {3x} \right)} \right]\frac{1}{{{{\sin }^4}\left( {3x} \right)}}\,dx}}\\ & = \int{{9\sin \left( {3x} \right) - 2\frac{{{{\cos }^3}\left( {3x} \right)}}{{{{\sin }^4}\left( {3x} \right)}}\,dx}}\end{align*}\] Show Step 2

The first integral is simple enough to do without any extra work.

For the second integral again, think about how we would do that if it was a product instead of a quotient. In that case we would simply strip out a cosine.

\[ \int{{\left[ {9{{\sin }^5}\left( {3x} \right) - 2{{\cos }^3}\left( {3x} \right)} \right]{{\csc }^4}\left( {3x} \right)\,dx}} = \int{{9\sin \left( {3x} \right) - 2\frac{{{{\cos }^2}\left( {3x} \right)}}{{{{\sin }^4}\left( {3x} \right)}}\cos \left( {3x} \right)\,dx}}\] Show Step 3

For the second integral we can use the trig identity \({\sin ^2}\theta + {\cos ^2}\theta = 1\) to convert the remaining cosines to sines.

\[ \int{{\left[ {9{{\sin }^5}\left( {3x} \right) - 2{{\cos }^3}\left( {3x} \right)} \right]{{\csc }^4}\left( {3x} \right)\,dx}} = \int{{9\sin \left( {3x} \right)\,dx}} - 2\int{{\frac{{1 - {{\sin }^2}\left( {3x} \right)}}{{{{\sin }^4}\left( {3x} \right)}}\cos \left( {3x} \right)\,dx}}\] Show Step 4

Now we can use the substitution \(u = \sin \left( {3x} \right)\) to evaluate the second integral. The first integral doesn’t need any extra work.

\[\begin{align*}\int{{\left[ {9{{\sin }^5}\left( {3x} \right) - 2{{\cos }^3}\left( {3x} \right)} \right]{{\csc }^4}\left( {3x} \right)\,dx}} & = \int{{9\sin \left( {3x} \right)\,dx}} - \frac{2}{3}\int{{\frac{{1 - {u^2}}}{{{u^4}}}\,du}}\\ & = \int{{9\sin \left( {3x} \right)\,dx}} - \frac{2}{3}\int{{{u^{ - 4}} - {u^{ - 2}}\,du}}\\ & = - 3\cos \left( {3x} \right) - \frac{2}{3}\left( { - \frac{1}{3}{u^{ - 3}} + {u^{ - 1}}} \right) + c\end{align*}\]

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.

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Don’t forget to substitute back in for \(u\)!

\[\begin{align*}\int{{\left[ {9{{\sin }^5}\left( {3x} \right) - 2{{\cos }^3}\left( {3x} \right)} \right]{{\csc }^4}\left( {3x} \right)\,dx}} & = - 3\cos \left( {3x} \right) + \frac{2}{9}\frac{1}{{{{\sin }^3}\left( {3x} \right)}} - \frac{2}{3}\frac{1}{{\sin \left( {3x} \right)}} + c\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 3\cos \left( {3x} \right) + \frac{2}{9}{{\csc }^3}\left( {3x} \right) - \frac{2}{3}\csc \left( {3x} \right) + c}}\end{align*}\]