What is the integral of (sin(2x))^2 ?

The integral of (sin(2x))^2 is 1/2 (x-1/4 sin(4x))+C

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integral of (sin(2x))^2

\int (sin (2 x))^{2} d x

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

Solution steps

\int (sin (2 x))^{2} d x

Rewrite using trig identities

= \int 4 cos^{2} (x) sin^{2} (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int cos^{2} (x) sin^{2} (x) d x

Rewrite using trig identities

= 4 \cdot \int \frac{1 - cos ( 4 x )}{8} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \frac{1}{8} \cdot \int 1 - cos (4 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 \cdot \frac{1}{8} (\int 1 d x - \int cos (4 x) d x)

\int 1 d x = x

\int cos (4 x) d x = \frac{1}{4} sin (4 x)

= 4 \cdot \frac{1}{8} (x - \frac{1}{4} sin (4 x))

Simplify

4 \cdot \frac{1}{8} (x - \frac{1}{4} sin (4 x)) : \frac{1}{2} (x - \frac{1}{4} sin (4 x))

= \frac{1}{2} (x - \frac{1}{4} sin (4 x))

Add a constant to the solution

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

\int r sin (r^{2}) d r

\frac{d y}{d t} = y + 2 e^{t}

\frac{\partial}{\partial x} (x^{2} + x y + y^{2})

\int \frac{2 x ^{2} + 6}{( x ^{2} - 2 x + 2 ) ^{2}} d x

x \to \infty lim (ln (2 x))

What is the integral of (sin(2x))^2 ?
The integral of (sin(2x))^2 is 1/2 (x-1/4 sin(4x))+C