What is the integral of 1/4 cos^2(2x) ?

The integral of 1/4 cos^2(2x) is 1/8 (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 1/4 cos^2(2x)

\int \frac{1}{4} cos^{2} (2 x) d x

\frac{1}{8} (x + \frac{1}{4} sin (4 x)) + C

Solution steps

\int \frac{1}{4} cos^{2} (2 x) d x

Take the constant out:

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

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

Rewrite using trig identities

= \frac{1}{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

= \frac{1}{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

= \frac{1}{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)

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{8} (x + \frac{1}{4} sin (4 x)) + C

y^{^{'}} + 0.2 \cdot y = 5

f (x) = x^{\frac{1}{6}}

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

\int t 4 + 9 t^{2} d t

\int \frac{1}{1 + 64 x ^{2}} d x

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