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

Question

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.

Answer 1

Your answer is correct.

Note that by using a different integration method you can get an answer which looks different but it is not.

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 $u=\cos(2x)$.