Calculus II - Integrals Involving Trig Functions

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Section 7.2 : Integrals Involving Trig Functions

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4. Evaluate \( \displaystyle \int_{\pi }^{{2\pi }}{{{{\cos }^3}\left( {\frac{1}{2}w} \right){{\sin }^5}\left( {\frac{1}{2}w} \right)\,dw}}\).

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Start Solution

We have two options for dealing with the limits. We can deal with the limits as we do the integral or we can evaluate the indefinite integral and take care of the limits in the last step. We’ll use the latter method of dealing with the limits for this problem.

In this case notice that both exponents are odd. This means that we can either strip out a cosine and convert the rest to sines or strip out a sine and convert the rest to cosines. Either are perfectly acceptable solutions. However, the exponent on the cosine is smaller and so there will be less conversion work if we strip out a cosine and convert the remaining cosines to sines.

Here is that work.

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

We can now use the substitution \(u = \sin \left( {\frac{1}{2}w} \right)\) to evaluate the integral.

\[\begin{align*}\int{{{{\cos }^3}\left( {\frac{1}{2}w} \right){{\sin }^5}\left( {\frac{1}{2}w} \right)\,dw}} & = 2\int{{\left( {1 - {u^2}} \right){u^5}\,du}}\\ & = 2\int{{{u^5} - {u^7}\,du}} = 2\left( {\frac{1}{6}{u^6} - \frac{1}{8}{u^8}} \right) + c\end{align*}\]

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 3

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

\[\int{{{{\cos }^3}\left( {\frac{1}{2}w} \right){{\sin }^5}\left( {\frac{1}{2}w} \right)\,dw}} = \frac{1}{3}{\sin ^6}\left( {\frac{1}{2}w} \right) - \frac{1}{4}{\sin ^8}\left( {\frac{1}{2}w} \right) + c\] Show Step 4

Now all we need to do is deal with the limits.

\[ \int_{\pi }^{{2\pi }}{{{{\cos }^3}\left( {\frac{1}{2}w} \right){{\sin }^5}\left( {\frac{1}{2}w} \right)\,dw}} = \left. {\left( {\frac{1}{3}{{\sin }^6}\left( {\frac{1}{2}w} \right) - \frac{1}{4}{{\sin }^8}\left( {\frac{1}{2}w} \right)} \right)} \right|_\pi ^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{1}{{12}}}}\] Show Alternate Solution

As we noted above we could just have easily stripped out a sine and converted the rest to cosines if we’d wanted to. We’ll not put that work in here, but here is the indefinite integral that you should have gotten had you done it that way.

\[\int{{{{\cos }^3}\left( {\frac{1}{2}w} \right){{\sin }^5}\left( {\frac{1}{2}w} \right)\,dw}} = - \frac{1}{2}{\cos ^4}\left( {\frac{1}{2}w} \right) + \frac{2}{3}{\cos ^6}\left( {\frac{1}{2}w} \right) - \frac{1}{4}{\cos ^8}\left( {\frac{1}{2}w} \right) + c\]

Note as well that regardless of which approach we use to doing the indefinite integral the value of the definite integral will be the same.