How to evaluate $\int \sin^2 x \ dx$

Question

The fact that sin is squared is really throwing me off, can't seem to relate it to any standard integrals.

Answer 1

HINT Remember that $\sin^2 x = 1 - \cos^2 x$ and use the double angle formula.

Answer 2

Let $$I=\int \sin^2(x)dx$$ and $$J=\int \cos^2(x)dx$$

$$I+J=\int dx=x+C_1$$ $$J-I=\int \cos(2x)dx=\frac{1}{2}\sin(2x)+C_2$$

thus by substraction

$$I=\frac{x}{2}-\frac{1}{4}\sin(2x)+C$$

Answer 3

$$\int\sin^2xdx\\=\int(1-\cos^2x)dx\\=\int dx-\int\cos^2x dx\\=x-\frac12\int(1+\cos2x)dx\\=x-\frac12x-\frac12\int\cos2x\ dx\\=\frac12x-\frac14\sin2x+C$$