Calculus II - 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 viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe 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

Back to Problem List

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

Show All Steps Hide All Steps

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.

Show Step 4

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

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