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

The integral of 2sin^2(x)cos^2(x) is 1/4 (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 2sin^2(x)cos^2(x)

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

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

Solution steps

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

Rewrite using trig identities

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

Simplify

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

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

Add a constant to the solution

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

((x + 5) e^{\frac{1}{x}})^{^{'}}

y = 6 x (2 x + 1)^{2} + (2 x + 1)^{3}

y^{^{''}} - 10 y^{^{'}} + 29 y = 0

\frac{d}{d x} (\frac{1}{x} + \frac{3}{x ^{2}} + \frac{2}{x ^{3}})

x \to \infty lim (co d s)

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