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.
Best Answer
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 $u=\cos(2x)$.