What is the integral of 1+cos(4θ) ?

The integral of 1+cos(4θ) is θ+1/4 sin(4θ)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1+cos(4θ)

\int 1 + cos (4 θ) d θ

θ + \frac{1}{4} sin (4 θ) + C

Solution steps

\int 1 + cos (4 θ) d θ

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d θ + \int cos (4 θ) d θ

\int 1 d θ = θ

\int cos (4 θ) d θ = \frac{1}{4} sin (4 θ)

= θ + \frac{1}{4} sin (4 θ)

Add a constant to the solution

= θ + \frac{1}{4} sin (4 θ) + C

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

\int \frac{4 y ^{2} - 6 y - 12}{y ( y + 2 ) ( y - 3 )} d y

\int e^{\frac{x}{x}} d x

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

\int (4 - 5 x) d x

What is the integral of 1+cos(4θ) ?
The integral of 1+cos(4θ) is θ+1/4 sin(4θ)+C