# Integral of $\sin^5(x)\cos(x)$

I'm trying to solve the following integral:

$$\int\sin^5(x)\cos(x)$$

I assumed I would do u-substitution where:

$$u = \sin(x)$$

$$du = \cos(x) dx$$

Which would then cancel out the $$\cos(x)$$

And leave me with:

$$\int u^5 du = \frac{u^6}{6} +C = \frac{\sin^6(x)}{6} + C$$

But apparently that is not correct?

Update: Seems it is the correct answer. The system I was using gave a different answer, so I plugged in a value into both the system's answer and my own answer, and got different results. Not sure why, but you can consider this closed then.

• $+C$ you forgot – AHusain Oct 15 '18 at 0:21
• What do you mean, "it's not correct?" Did you follow all the instructions your online homework system (coughcough Webassign) gave you? – Sean Roberson Oct 15 '18 at 0:24
• This is fully correct. Actually no substitution is needed, because this is in the form $\int g'(f(x))f'(x)\,dx=g(f(x))+c$, with $g(x)=x^6/6$ and $f(x)=\sin x$. – egreg Oct 15 '18 at 0:24
• @SeanRoberson So the answer they gave is in a different form, and I tried plugging in a value into their answer, and a value into my answer, and I got different results. So I assumed they weren't the same but I guess they are. – James Mitchell Oct 15 '18 at 0:29
• If differentiating the two solutions gives back $\sin^5x\cos x$, then both are correct. – egreg Oct 15 '18 at 0:31

For example $$\int\sin^5(x)\cos(x)=\int\sin(x) (1-\cos^2(x))^2 \cos(x) dx$$can becalculayted using the substitution $$v=\cos(x)$$. If you do this, the answer loos different, but that's just an illusion.
Same way, you can use $$\sin^5(x)\cos(x)=\left(\frac{1-\cos(2x)}{2}\right)^2\frac{\sin(2x)}{2}$$ and then the substitution $$u=\cos(2x)$$.