**HOMEWORK SPOILER ALERT**
On the latest homework assignment, the trickiest one for me was the following:
\int_\ 160cos^4(20x)dx
I eventually found a solution, but it was a really long process:
\int_\ 160cos^4(20x)dx = 160 \int_\ cos^4(20x)dx
I did a little u- substitution:
\mathrm{let}\ u = 20x
\frac{du}{dx} = 20
\frac{dx}{du} = \frac{1}{20}
dx = \frac{1}{20} du
160 \int_\ cos^4(20x)dx = 160 \int_\ cos^4(u) \frac{1}{20} du
= 8 \int_\ cos^4(u)du
Then I used a half-angle formula:
8 \int_\ cos^4(u)du = 8 \int_\ (cos^2(u))^2du = 8 \int_\ (\frac{1}{2} (1 + cos(2u)))^2du
= 8 \int_\ \frac{1}{4} (1 + cos(2u))^2du
= 2 \int_\ (1 + cos(2u))^2du
Then I did another u- substitution, only I couldn’t use u this time:
\mathrm{let}\ q = 2u
\frac{dq}{du} = 2
\frac{du}{dq} = \frac{1}{2}
du = \frac{1}{2} dq
2 \int_\ (1 + cos(2u))^2du = 2 \int_\ (1 + cos(q))^2 * \frac{1}{2}dq
= \int_\ (1 + cos(q))^2 dq
I expanded (1 + cos(q))^2:
\int_\ (1 + cos(q))^2 dq = \int_\ (1 + 2cos(q) +cos^2(q)) dq
And then used the same half-angle formula again:
\int_\ (1 + 2cos(q) +cos^2(q)) dq = \int_\ (1 + 2cos(q) +\frac{1}{2}(1+cos(2q)))dq
And then converted that integral into the sum of two integrals:
\int_\ (1 + 2cos(q) +\frac{1}{2}(1+cos(2q)))dq = \int_\ (1 + 2cos(q))dq + \int_\ \frac{1}{2}(1+cos(2q))dq
= \int_\ (1 + 2cos(q))dq + \frac{1}{2}\int_\ (1+cos(2q))dq
I integrated the first part:
\int_\ (1 + 2cos(q))dq = q + 2sin(q)
And then did yet another u- substitution on the second part:
\mathrm{let}\ r = 2q
\frac{dr}{dq} = 2
\frac{dq}{dr} = \frac{1}{2}
dq = \frac{1}{2}dr
\int_\ (1+cos(2q))dq = \int_\ (1+cos(r))\frac{1}{2}dr
I could finally finish integrating:
\int_\ (1+cos(r))\frac{1}{2}dr
= \frac{1}{2}\int_\ (1+cos(r))dr
= \frac{1}{2}(r+sin(r))
And then translated r back into q and q back into u and u back into x:
q + 2sin(q) +(\frac{1}{2}*\frac{1}{2}(r+sin(r)))
= q + 2sin(q) +\frac{1}{4}(r+sin(r))
= q + 2sin(q) +\frac{1}{4}(2q+sin(2q))
= 2u + 2sin(2u) +\frac{1}{4}(2(2u)+sin(2(2u)))
= 2u + 2sin(2u) +\frac{1}{4}(4u + sin(4u))
= 2u + 2sin(2u) +\frac{1}{4}4u + \frac{1}{4}sin(4u)
= 2u + 2sin(2u) + u + \frac{sin(4u)}{4}
= 3u + 2sin(2u) + \frac{sin(4u)}{4}
= 3(20x) + 2sin(2(20x)) + \frac{sin(4(20x))}{4}
= 60x + 2sin(40x) + \frac{sin(80x)}{4}
\ \ \ \ \ \ This was the correct answer, but the method was a monstrosity of both length and sheer ugliness. I shudder to imagine how long this kind of process would have taken if the integral used cos^8 instead of cos^4. \ My question to fellow students: Did you - or could you - solve this integral in a more elegant ^* but equally rigorous way?
^* By ‘more elegant’ I don’t mean ‘using more mental math’. I mean actually simpler and more efficient.
