What is the correct way to solve the integral of cos^5 x sin x?

[DaleSwanson]

When I did this problem I overlooked the fact that a u-sub of u=cos x would work fine. I ended up using the more complicated trig integrals rules and did this:
∫cos⁵ x sin x dx
∫(cos² x)² cos x sin x dx
∫(1 - sin² x)² cos x sin x dx
∫(sin⁵ x - 2 sin³ + sin x) cos x dx
with u = sin x, du = cos x dx this is a simple integral. However it gives me:
$\frac{1}{6} sin^{6} x - \frac{1}{2} sin^4 x \frac{1}{2} sin^2 x$
The correct answer is:
$\frac{-cos^{6} x}{6}$

Wolfram Alpha confirms that the final form of the integral I gave above still gives the correct answer. It also confirms that my answer isn't numerically equivalent to the correct answer. Also that my integration of the u-subed version is correct. Therefore, I must assume there was some mistake in the u-sub itself, either in the initial sub or the replacement after the integration. I cannot see where this mistake is though.

Just to be clear, I get the correct answer when I use the simpler u-sub of u = cos x, and I understand that is the preferred way to go about this problem. However, as far as I can tell the more complicated way I did it should have still produced a correct answer.

 

[D H]

When I did this problem I overlooked the fact that a u-sub of u=cos x would work fine. I ended up using the more complicated trig integrals rules and did this:
∫cos⁵ x sin x dx
∫(cos² x)² cos x sin x dx

You could have used a u substitution here as well with u=cos²x, du=-2cos(x)*sin(x) dx.

∫(1 - sin² x)² cos x sin x dx

And here with u=sin²x, du=2cos(x)*sin(x) dx.

∫(sin⁵ x - 2 sin³ + sin x) cos x dx
with u = sin x, du = cos x dx this is a simple integral. However it gives me:
$\frac{1}{6} sin^{6} x - \frac{1}{2} sin^4 x \frac{1}{2} sin^2 x$

I assume this last line is a typo and that you meant $\frac{1}{6} sin^{6} x - \frac{1}{2} sin^4 x + \frac{1}{2} sin^2 x$.

The correct answer is:
$\frac{-cos^{6} x}{6}$

Your more complicated answer is also correct.

You are forgetting the constant of integration. Your complex answer and the simple one differ by a constant.

 

[DaleSwanson]

Your more complicated answer is also correct.

You are forgetting the constant of integration. Your complex answer and the simple one differ by a constant.

That possibility did occur to me briefly, but for some reason I quickly dismissed it. Thanks.

 

[D H]

That possibility did occur to me briefly, but for some reason I quickly dismissed it. Thanks.

That constant of integration can always be written as c*1, but where "1" is written in an interesting way. A good start is to represent 1 as sin²x+cos²x. Cube this and simplify a bit and you will find a way to express one that applies to this problem, sin⁶x-3sin⁴x+3sin²x+cos⁶x.