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

The integral of 2(sin(2x))^2 is 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 2(sin(2x))^2

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

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

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

= 2 \cdot \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

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

Rewrite using trig identities

= 2 \cdot 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

= 2 \cdot 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

= 2 \cdot 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)

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (x y \cdot e^{x y})

\frac{d}{d x} (3 x^{2} - e^{x})

\int 4^{t} sin (4^{t}) d t

\frac{d}{d x} (y - x + 1 + x - y + 1)

f (x) = ln (x + 9)

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