Substitutions for the trigonometric integral $\int \cos \theta \cos^5\left(\sin \theta\right) d\theta$

Question

I evaluated $$\int \cos \theta \cos^5\left(\sin \theta\right) d\theta$$ and got to a result that is very similar to what I can check with Mathematica but I am not sure if it is equivalent, what I did is the following, $$\int \cos \theta \cos^5\left(\sin \theta\right) d\theta$$ $$= \int \cos^4\left(\sin \theta\right)\cos\left(\sin\theta\right)\cos \theta d\theta$$ $$= \frac 14\int \left(1+2\cos\left(2\sin\theta\right)+\cos^2\left(2\sin\theta\right)\right)\cos\theta d\theta$$ $$=\frac 14 \int \cos\left(\sin\theta\right)\cos \theta d\theta\;+\;\frac 12 \int \cos\left(2\sin\theta\right)\cos\left(\sin\theta\right)\cos \theta d\theta\;$$ $$+\;\frac 18 \int \left(1+\cos\left(4\sin \theta\right)\right)\cos\left(\sin\theta\right)\cos\theta d\theta$$ I use $\cos A \cos B = \frac 12 \left(\cos\left(A-B\right) + \cos\left(A+B\right)\right)$ and I get, $$= \frac 14 \int \cos\left(\sin\theta\right)\cos\theta d\theta\;+\;\frac 14 \int \cos\left(\sin \theta\right)\cos\theta\,d\theta\,+\, \frac 14 \int\cos\left(3\sin\theta\right)\cos\theta d\theta \;$$ $$+\; \frac 1{8} \int \cos\left(\sin\theta\right)\cos \theta\,d\theta \;+\; \frac 1{16} \int \cos\left(3\sin\theta\right)\cos \theta d\theta \;+\;\frac 1{16}\int \cos\left(5\sin\theta\right)\cos \theta d\theta$$ wihch is equal to $$=\frac 5{8} \int \cos\left(\sin\theta\right)\cos \theta\,d\theta \;+\;\frac 5{16} \int \cos\left(3\sin\theta\right)\cos \theta d\theta \;+\;\frac 1{16}\int \cos\left(5\sin\theta\right)\cos \theta d\theta$$

Edit (correct solution)

Applying the correct substitution, letting $u = \sin \theta$ such that $du = \cos \theta\,d\theta$ to get, $$\frac 58 \int \cos u\,du\;+\;\frac 5{16} \int \cos 3u\, du\;+\; \frac 1{16} \int \cos 5u \,du$$ yields the same result as $Mathematica$, that is, $$\int \cos \theta \cos^5\left(\sin \theta\right) d\theta \;=\; \frac 58 \sin\left(\sin\theta\right)\;+\; \frac 5{48} \sin\left(3\sin\theta\right) \;+\; \frac 1{80} \sin\left(5\sin\theta\right)$$ Of course the substitution could and should be applied right at the beginning, for the sake of simplicity. I wanted to correct the last step while leaving the question as it was for others that might make the same mistake I did and for whomever might find that information useful.

Answer 1

Start by organizing things a little better: we have $$ \int \cos^5(\sin(\theta))\cos(\theta)d\theta = \int \cos^5(u)du $$ where $u=\sin(\theta).$ Thus we only need to worry about this integral.

One way to do this is to write $$ \int \cos^5(u)du = \int\cos^4(u)\cos(u)du = \int(1-\sin^2(u))^2\cos(u)du = \int(1-w^2)^2dw$$ where $w=\sin(u).$ Thus we get $$ \int(1-w^2)^2dw = \int(1-2w^2+w^4)dw = w-\frac{2}{3}w^3 + \frac{1}{5}w^4 \\= \sin(\sin(\theta)) - \frac{2}{3}\sin^3(\sin(\theta))+\frac{1}{5}\sin^5(\sin(\theta))$$ where we plugged back in $w=\sin(\sin(\theta)).$

Another way is to use a formula for $\cos^5(u)$: $$\int \cos^5(u)du = \int\frac{1}{16}(10\cos(u)+5\cos(3u)+\cos(5u))du\\=\frac{1}{16}\left(10\sin(u)+\frac{5}{3}\sin(3u) + \frac{1}{5}\sin(5u) \right)\\=\frac{5}{8}\sin(\sin(\theta)) + \frac{5}{48}\sin(3\sin(\theta)) + \frac{1}{80}\sin(5\sin(\theta)).$$ Apparently Mathematica likes this way.

So it's entirely possible to get very different looking expressions. Yours is not right, however. While you do a lot of correct stuff in the simplification phase, you are right that where you say "this is when I'm not sure I'm correct" is where you go off the rails. You should perform the u substituation $u=\sin(\theta)$ to get rid of the $\cos(\theta),$ not use a trig identity to absorb it (which you didn't do right.. you should get things like $\cos(\sin(\theta)-\theta))$... you can't subtract bring the $\theta$ inside the inner $\sin$ function. It doesn't work that way.)

Then when you integrate, that's not right either. You have for one term $\int \cos(\sin(2\theta))d\theta .$ You cannot simply integrate this by dividing by what's inside. This only words when the inside is $3\theta$ or something like that... then you can divide by $3.$ Instead you would need something like a $\cos(2\theta)$ on the outside that you could use to u substitute $u=\sin(2\theta),$ but you don't have that (and the expression isn't right, anyhow).

I see you're using Mathematica to check your answers, and this is a good idea (provided you don't become too reliant on it). A couple tips:

1. If you're wondering if two expressions are equivalent, you can use mathematica to plot them and see.

2. You can use mathematica to check substeps as well as the final answer, as in $\int \sin(\sin(2\theta))d\theta$ above.

Answer 2

Try this method once -

Put $\sin\theta =t$

$\cos\theta d\theta = dt$

So we have,

$\int \cos^5t$ dt

Now solve it.

Any doubt feel free to ask.