Confused about washers

Question

I am given the equation $\sin x$ and I'm to rotate the interval $(0,\pi)$ around the $x$-axis. I think this is equal to $$\int_0^\pi \pi\left(1^2-(\sin x)^2\right)\ dx.$$ Am I correct? If not, where did I go wrong? Can you also help me find the indefinite integral of $(\sin x)^2$

Answer 1

If the region is between $y=\sin x$ and the $x$-axis, then you should have disks of radius $\sin^2 x$; your integral is for revolving the region between $y=\sin x$ and $y=1$ about the $x$-axis.

To integrate $\sin^2x$ and $\cos^2x$ you use the half-angle formulas:

$$\sin^2x=\frac{1-\cos 2x}2\quad\text{and}\quad\cos^2x=\frac{1+\cos 2x}2\;.\tag{1}$$

These are both deduced from the double angle formula $\cos 2x=\cos^2x-\sin^2x$ and the Pythagorean identity $\sin^2x+\cos^2x=1$. For example, to get the first formula in $(1)$, start with

$$\cos 2x=\cos^2x-\sin^2x=\left(1-\sin^2x\right)-\sin^2x=1-2\sin^2x$$

and solve for $\sin^2x$.