<s>[INST] Substitutions for the trigonometric integral $\int \cos \theta \cos^5\left(\sin \theta\right) d\theta$ 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.
 [/INST] 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.
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