What is the integral of (1/2 (1-cos(2x)))^2 ?

The integral of (1/2 (1-cos(2x)))^2 is -1/4 sin^3(x)cos(x)+3/8 (x-1/2 sin(2x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1/2 (1-cos(2x)))^2

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

- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)) + C

Solution steps

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

Simplify

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

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

Apply Integral Reduction

= - \frac{cos ( x ) sin ^{3} ( x )}{4} + \frac{3}{4} \cdot \int sin^{2} (x) d x

\int sin^{2} (x) d x = \frac{1}{2} (x - \frac{1}{2} sin (2 x))

= - \frac{cos ( x ) sin ^{3} ( x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (x - \frac{1}{2} sin (2 x))

Simplify

- \frac{cos ( x ) sin ^{3} ( x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (x - \frac{1}{2} sin (2 x)) : - \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x))

= - \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x))

Add a constant to the solution

= - \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)) + C

x \to - 0.1 lim (x^{3} + 6 x^{2} + x)

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

\int \frac{3 x ^{2} - 8 x + 1}{x ^{3} - 4 x ^{2} + x + 6} d x

\int sin (2 x) sin (4 x) sin (6 x) d x

\int \frac{x ^{\frac{3}{2}} - 2 x ^{6} - 1.2}{x} d x

What is the integral of (1/2 (1-cos(2x)))^2 ?
The integral of (1/2 (1-cos(2x)))^2 is -1/4 sin^3(x)cos(x)+3/8 (x-1/2 sin(2x))+C