Calculus II - Integrals Involving Trig Functions

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Section 7.2 : Integrals Involving Trig Functions

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2. Evaluate \( \displaystyle \int{{{{\sin }^8}\left( {3z} \right){{\cos }^5}\left( {3z} \right)\,dz}}\) .

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Start Solution

The first thing to notice here is that the exponent on the cosine is odd and so we can strip one of them out.

\[ \int{{{{\sin }^8}\left( {3z} \right){{\cos }^5}\left( {3z} \right)\,dz}} = \int{{{{\sin }^8}\left( {3z} \right){{\cos }^4}\left( {3z} \right)\cos \left( {3z} \right)\,dz}}\] Show Step 2

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

\[\begin{align*}\int{{{{\sin }^8}\left( {3z} \right){{\cos }^5}\left( {3z} \right)\,dz}} & = \int{{{{\sin }^8}\left( {3z} \right){{\left[ {{{\cos }^2}\left( {3z} \right)} \right]}^2}\cos \left( {3z} \right)\,dz}}\\ & = \int{{{{\sin }^8}\left( {3z} \right){{\left[ {1 - {{\sin }^2}\left( {3z} \right)} \right]}^2}\cos \left( {3z} \right)\,dz}}\end{align*}\] Show Step 3

We can now use the substitution \(u = \sin \left( {3z} \right)\) to evaluate the integral.

\[\begin{align*}\int{{{{\sin }^8}\left( {3z} \right){{\cos }^5}\left( {3z} \right)\,dz}} & = \frac{1}{3}\int{{{u^8}{{\left[ {1 - {u^2}} \right]}^2}\,du}}\\ & = \frac{1}{3}\int{{{u^8} - 2{u^{10}} + {u^{12}}\,du = \frac{1}{3}\left( {\frac{1}{9}{u^9} - \frac{2}{{11}}{u^{11}} + \frac{1}{{13}}{u^{13}}} \right) + c}}\end{align*}\]

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 \(u\)!

\[ \int{{{{\sin }^8}\left( {3z} \right){{\cos }^5}\left( {3z} \right)\,dz}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{27}}{{\sin }^9}\left( {3z} \right) - \frac{2}{{33}}{{\sin }^{11}}\left( {3z} \right) + \frac{1}{{39}}{{\sin }^{13}}\left( {3z} \right) + c}}\]