What is the integral of 4sin^2(2x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 4sin^2(2x)

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

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

Solution steps

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

Rewrite using trig identities

= 4 \cdot \int \frac{1}{2} (1 - cos (4 x)) d x

Take the constant out:

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{( x ^{2} - 2 x )}{x + 1})

\frac{d y}{d x} = e^{y} cos (2 x)

f (t) = 2 cos (t)

x = - 2, x = 1, y = 3 x, y = x^{2} - 4

y^{^{''}} + y = 5 csc^{2} (t)

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