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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.

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  • $\begingroup$ Why don't you use mathematica to plot both of those functions and see if they are the same (or differ by a constant)? $\endgroup$ – spaceisdarkgreen Feb 13 '17 at 2:16
  • $\begingroup$ I can guarantee that your answer is not correct because the integrand is always continuous so the anti-derivative should also be continuous everywhere. Although you might be able to simplify your expression to get the one from mathematica. $\endgroup$ – lordoftheshadows Feb 13 '17 at 2:17
  • $\begingroup$ @lordoftheshadows where is the antiderivative not continuous? (and if it weren't they wouldn't be able to simplify the expression to get the correct expression) $\endgroup$ – spaceisdarkgreen Feb 13 '17 at 2:18
  • $\begingroup$ $\theta = 0$. You have a removable discontinuity which is why I suspect that it may just be an algebra problem. $\endgroup$ – lordoftheshadows Feb 13 '17 at 2:19
  • $\begingroup$ I graphed it and those two functions don't look the same. Try $\theta = -\pi/2$ $\endgroup$ – lordoftheshadows Feb 13 '17 at 2:22
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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.

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  • $\begingroup$ Oh! I did see the substitution in the first place and I do not know why I thought it was not allowed in that particular case, so I tried to crack it without substitution. Indeed I get to the same result when using the correct substitution as you did. I overcomplicated things way too much here, anyway we also learn by making errors right? :) thank you for the trig and $Mathematica$ tips and for taking the time. $\endgroup$ – user409521 Feb 13 '17 at 2:56
  • $\begingroup$ @NoBoundaries Yes, I agree with the old quote "an expert is someone who has made every mistake one can make in a field" $\endgroup$ – spaceisdarkgreen Feb 13 '17 at 3:01
  • $\begingroup$ Well agreed and at the same time, you answered a few trig questions I had been wondering about for a while so in the end it really was a very useful mistake! $\endgroup$ – user409521 Feb 13 '17 at 3:04
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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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