5 Solutions to Problem 7.2.10

Problem 7.2.10

This took at least two hours to put together and write up. Just letting
you know the effort I'm willing to put in to help you do well.

Solution 1:

Some students used integration by parts. Here's my version of their
method:

\begin{alignat*}{1}
A & =\int_{0}^{\pi}\sin^{2}(t)\cos^{4}(t)dt\\
& \begin{cases}
u=\sin^{2}(t)\cos^{3}(t) & dv=\cos(t)dt\\
du=2\sin(t)\cos^{4}(t)-3\sin^{3}(t)\cos^{2}(t) & v=\sin(t)
\end{cases}\\
& =\left.\sin^{2}(t)\cos^{3}(t)\sin(t)\right|_{0}^{\pi}-\int_{0}^{\pi}2\sin^{2}(t)\cos^{4}(t)-3\sin^{4}(t)\cos^{2}(t)dt\\
& =0-0-2f(t)+3\int_{0}^{\pi}\sin^{4}(t)\cos^{2}(t)dt\\
3A & =3\int_{0}^{\pi}\sin^{4}(t)\cos^{2}(t)dt\\
A & =\int_{0}^{\pi}\sin^{4}(t)\cos^{2}(t)dt\\
& =\int_{0}^{\pi}(1-\cos^{2}(t))\sin^{2}(t)\cos^{2}(t)dt\\
& =\int_{0}^{\pi}\sin^{2}(t)\cos^{2}(t)dt-A\\
2A & =\int_{0}^{\pi}\sin^{2}(t)\cos^{2}(t)dt\\
& =\int_{0}^{\pi}\frac{1-\cos2t}{2}\cdot\frac{1+\cos2t}{2}dt\\
& =\frac{1}{4}\int_{0}^{\pi}1-\cos^{2}2tdt\\
& =\frac{1}{4}\int_{0}^{\pi}1-\frac{1+\cos4t}{2}dt\\
& =\frac{1}{4}\int_{0}^{\pi}\frac{1}{2}-\frac{\cos4t}{2}dt\\
& =\frac{1}{4}\left.\left(\frac{t}{2}-\frac{\sin4t}{8}\right)\right|_{0}^{\pi}\\
& =\frac{\pi}{8}\\
A & =\frac{\pi}{16}
\end{alignat*}

Note: I found a student used the curly brace $\{$ between a sequence
of equalities had such an elegant effect that I've decided to adopt
the usage of the curly brace in such situations. Thanks!

Solution 2:

Various students used the reduction formula for cosine.
\[
\int_{0}^{\pi}\cos^{n}(\theta)d\theta=\left.\frac{1}{n}\cos^{n-1}(\theta)\sin(\theta)\right|_{0}^{\pi}+\frac{n-1}{n}\int_{0}^{\pi}\cos^{n-2}(\theta)d\theta
\]
However, they used it inefficiently. Here's my efficient use of the
formula.

\begin{alignat*}{1}
\int_{0}^{\pi}\sin^{2}(t)\cos^{4}(t)dt & =\int_{0}^{\pi}(1-\cos^{2}(t))\cos^{4}(t)dt\\
& =\int_{0}^{\pi}\cos^{4}(t)dt-\int_{0}^{\pi}\cos^{6}(t)dt\\
& =\int_{0}^{\pi}\cos^{4}(t)dt-\left[0-0+\frac{5}{6}\int_{0}^{\pi}\cos^{4}(t)dt\right]\\
& =\frac{1}{6}\int_{0}^{\pi}\cos^{4}(t)dt\\
& =\frac{1}{6}\left[0-0+\frac{3}{4}\int_{0}^{\pi}\cos^{2}(t)dt\right]\\
& =\frac{1}{8}\int_{0}^{\pi}\cos^{2}(t)dt\\
& =\frac{1}{8}\left[0-0+\frac{1}{2}\int_{0}^{\pi}dt\right]\\
& =\frac{1}{16}\left.t\vphantom{\frac{}{}}\right|_{0}^{\pi}=\frac{\pi}{16}
\end{alignat*}

Solution 3A:

Various students used the strategy in the book for when both the powers
of sine and cosine are even. That is, use the identites
\[
\cos^{2}(\theta)=\frac{1+\cos(2\theta)}{2}
\]
\[
\sin^{2}(\theta)=\frac{1-\cos(2\theta)}{2}
\]

\begin{alignat*}{1}
\int_{0}^{\pi}\sin^{2}(t)\cos^{4}(t)dt & =\int_{0}^{\pi}\left(\frac{1-\cos(2t)}{2}\right)\left(\frac{1+\cos(2t)}{2}\right)^{2}dt\\
& =\frac{1}{8}\int_{0}^{\pi}\left(1-\cos^{2}(2t)\right)(1+\cos(2t))dt\\
& =\frac{1}{8}\int_{0}^{\pi}1+\cos(2t)-\cos^{2}(2t)-\cos^{3}(2t)dt\\
& =\frac{1}{8}\int_{0}^{\pi}1+\cos(2t)-\frac{1+\cos(4t)}{2}-(1-\sin^{2}(2t))\cos(2t)dt\\
& =\frac{1}{8}\int_{0}^{\pi}\frac{1}{2}-\frac{1}{2}\cos(4t)+\sin^{2}(2t)\cos(2t)dt\\
& =\frac{1}{8}\left.\left(\frac{t}{2}-\frac{1}{8}\sin(4t)+\frac{1}{6}\sin^{3}(2t)\right)\right|_{0}^{\pi}\\
& =\frac{1}{8}\left[\left(\frac{\pi}{2}-0+0\right)-\left(\vphantom{\frac{}{}}0-0+0\right)\right]\\
& =\frac{\pi}{16}
\end{alignat*}

Remark: A common mistake is to attempt the $u$-substitution $u=\cos(2t)$
after getting to $\frac{1}{8}\int_{0}^{\pi}1+\cos(2t)-\cos^{2}(2t)-\cos^{3}(2t)dt$.
Unfortunately, we should remember we would also have $du=-2\sin(2t)dt$,
so we cannot make good use of the $u$-substitution.

Solution 3B:

Note that I wanted to write out the solution above, to show the methods
used to find the indefinite integral prior to evaluating the definite
integral. However, if I wasn't interested in finding the antiderivate
(indefinite integral), then I could take advantage of the domain of
integration. We can argue that
\[
\int_{0}^{\pi}\cos^{2k+1}(nt)dt=0\text{ for }k\geq0
\]
without doing computition. However, I'm treading into unsafe territory
and simply want to bring the possibility to your attention. In the
end I advise you stick to Solution 3A and 4A.

Assuming what I wrote above is true, we have

\begin{alignat*}{1}
\int_{0}^{\pi}\sin^{2}(t)\cos^{4}(t)dt & =\int_{0}^{\pi}\left(\frac{1-\cos(2t)}{2}\right)\left(\frac{1+\cos(2t)}{2}\right)^{2}dt\\
& =\frac{1}{8}\int_{0}^{\pi}\left(1-\cos^{2}(2t)\right)(1+\cos(2t))dt\\
& =\frac{1}{8}\int_{0}^{\pi}1+\cos(2t)-\cos^{2}(2t)-\cos^{3}(2t)dt\\
& =\frac{1}{8}\int_{0}^{\pi}1-\cos^{2}(2t)dt\\
& =\frac{1}{8}\int_{0}^{\pi}1-\frac{1+\cos(4t)}{2}dt\\
& =\frac{1}{8}\int_{0}^{\pi}\frac{1}{2}-\frac{1}{2}\cos(4t)dt\\
& =\frac{1}{8}\int_{0}^{\pi}\frac{1}{2}dt\\
& =\frac{\pi}{16}
\end{alignat*}

Solution 4A:

Some students used the identity $2\sin\theta\cos\theta=\sin2\theta$
at the beginning of the problem. After that, there are various ways
to deal with the result. Here is one possible way to proceed. (Note
that this method is not so different from Solution 3A. Can you find
all the places where you can jump from one solution to the other?)

\begin{alignat*}{1}
\int_{0}^{\pi}\sin^{2}(t)\cos^{4}(t)dt & =\int_{0}^{\pi}\left(\frac{\sin(2t)}{2}\right)^{2}\cos^{2}(t)dt\\
& =\int_{0}^{\pi}\left(\frac{\sin(2t)}{2}\right)^{2}\left(\frac{1+\cos(2t)}{2}\right)dt\\
& =\frac{1}{8}\int_{0}^{\pi}\sin^{2}(2t)+\sin^{2}(2t)\cos(2t)dt\\
& =\frac{1}{8}\int_{0}^{\pi}\frac{1-\cos(4t)}{2}+\sin^{2}(2t)\cos(2t)dt\\
& =\frac{1}{8}\left.\left(\frac{t}{2}-\frac{1}{8}\sin(4t)+\frac{1}{6}\sin^{3}(2t)\right)\right|_{0}^{\pi}\\
& =\frac{1}{8}\left[\left(\frac{\pi}{2}-0+0\right)-\left(\vphantom{\frac{}{}}0-0+0\right)\right]\\
& =\frac{\pi}{16}
\end{alignat*}

Solution 4B:

I won't write out this in full, but like 3B, we want to take advantage
of the domain. We try to argue that $\int_{0}^{\pi}\sin^{2}(2t)\cos(2t)dt=0$
without needing to compute it. However, I'm treading into unsafe territory
and simply want to bring the possibility to your attention. In the
end I advise you stick to Solution 3A and 4A.

Solution 5:

Similar start to solution 4.

\begin{alignat*}{1}
\int_{0}^{\pi}\sin^{2}(t)\cos^{4}(t)dt & =\int_{0}^{\pi}\left(\frac{\sin(2t)}{2}\right)^{2}\cos^{2}(t)dt\\
& =\int_{0}^{\pi}\left(\frac{\sin(2t)}{2}\right)^{2}\left(\frac{1+\cos(2t)}{2}\right)dt\\
& =\frac{1}{8}\int_{0}^{\pi}\left(\frac{1-\cos(4t)}{2}\right)\left(1+\cos(2t)\right)dt\\
& =\frac{1}{16}\int_{0}^{\pi}1+\cos(2t)-\cos(4t)-\cos(4t)\cos(2t)dt\\
& =\frac{1}{16}\int_{0}^{\pi}1+\cos(2t)-\cos(4t)-\frac{1}{2}\left[\cos(2t)+\cos(6t)\right]dt\\
& =\frac{1}{16}\left.\left(t+\frac{1}{4}\sin(2t)-\frac{1}{4}\sin(4t)-\frac{1}{12}\sin(6t)\right)\right|_{0}^{\pi}\\
& =\frac{\pi}{16}
\end{alignat*}