What is the integral of 1+cos(2θ) ?

The integral of 1+cos(2θ) is θ+1/2 sin(2θ)+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(2θ)

\int 1 + cos (2 θ) d θ

θ + \frac{1}{2} sin (2 θ) + C

Solution steps

\int 1 + cos (2 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 (2 θ) d θ

\int 1 d θ = θ

\int cos (2 θ) d θ = \frac{1}{2} sin (2 θ)

= θ + \frac{1}{2} sin (2 θ)

Add a constant to the solution

= θ + \frac{1}{2} sin (2 θ) + C

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

\int \frac{x ^{3}}{( x + 1 ) ^{2} + 4} d x

1 6^{4} 4 x^{4} + 4

6 y^{^{'}} = x^{2}

\frac{d}{d x} (\frac{1}{3} x^{3} - x^{2} - 8 x + 1)

What is the integral of 1+cos(2θ) ?
The integral of 1+cos(2θ) is θ+1/2 sin(2θ)+C