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

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

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

\frac{3}{4} x + \frac{1}{4} sin (2 x) + sin (x) + C

Solution steps

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

Simplify

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

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

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

\int \frac{1}{4} d x = \frac{1}{4} x

\int cos (x) d x = sin (x)

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

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

Simplify

\frac{1}{4} x + sin (x) + \frac{1}{2} (x + \frac{1}{2} sin (2 x)) : \frac{3}{4} x + \frac{1}{4} sin (2 x) + sin (x)

= \frac{3}{4} x + \frac{1}{4} sin (2 x) + sin (x)

Add a constant to the solution

= \frac{3}{4} x + \frac{1}{4} sin (2 x) + sin (x) + C

\int (8 x^{2} + 16 x) (4 x^{3} + 12 x^{2})^{- 3} d x

\int sin (2 - 4 x) d x

\int \frac{- x + 2}{x ^{2} + 3} d x

\int + cos (2 n t) d t

\int (t + 1) e^{- t} d t

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