What is the integral of 2sin^2(x) ?

The integral of 2sin^2(x) is 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 2sin^2(x)

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

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

Solution steps

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

Take the constant out:

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

= 2 \cdot \int sin^{2} (x) d x

Rewrite using trig identities

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

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int 1 - 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

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

\int 1 d x = x

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

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

Simplify

2 \cdot \frac{1}{2} (x - \frac{1}{2} sin (2 x)) : x - \frac{1}{2} sin (2 x)

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

Add a constant to the solution

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

x^{3} + y^{2} x^{4} + 4 y^{^{'}} = 0

\int_{0}^{x} (e^{- t^{2}}) d t

x 1 + y^{2} d x = y 1 + x^{2} d y

\frac{d}{d x} (\frac{f ( x )}{g ( x ) - 2})

\frac{d y}{d x}, 3 x y + 1 = 0

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