What's the integration of $\int \sin^4 (x)dx$? What's the integration of $\int \sin^4 x \,dx$? I don't see the approach to this question.
I have a issue with this question as well:
$$\int \sin x \cos x (\sin x+\cos x) \,dx.$$
I simplify this to $\sin^2 x \cos x + \sin x \cos^2 x$
and set $u= \sin x$, $du = \cos x \,dx$.
So I got
$$\int u^2 \,du + ???.$$
I don't get the second part now

Recall the identities
$$\sin^2(\theta) = \dfrac{1-\cos(2\theta)}2$$
and
$$\cos^2(\theta) = \dfrac{1+\cos(2\theta)}2$$
Hence,
$$\sin^4(x) = \left(\dfrac{1-\cos(2x)}2 \right)^2 = \dfrac{1 - 2 \cos(2x) + \cos^2(2x)}4 = \dfrac{1 - 2 \cos(2x) + \dfrac{1+\cos(4x)}2}4$$
Hence,
$$\sin^4(x) = \dfrac{3-4\cos(2x) + \cos(4x)}8$$
Now you should be able to integrate this term by term and obtain an answer.
For the second part, note that $$\int \sin^2(x) \cos(x) dx = \int \sin^2(x) d(\sin(x))$$ and $$\int \cos^2(x) \sin(x) dx = -\int \cos^2(x) d(\cos(x))$$
