What is the integral of [cos(x/2)-sin(3 x/2)] ?

The integral of [cos(x/2)-sin(3 x/2)] is 2(sin(x/2)+1/3 cos((3x)/2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of [cos(x/2)-sin(3 x/2)]

\int [cos (\frac{x}{2}) - sin (3 \frac{x}{2})] d x

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

Solution steps

\int cos (\frac{x}{2}) - sin (3 \cdot \frac{x}{2}) d x

Apply u-substitution

: \int (cos (u) - sin (3 u)) \cdot 2 d u

= \int (cos (u) - sin (3 u)) \cdot 2 d u

Take the constant out:

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

= 2 \cdot \int cos (u) - sin (3 u) d u

Apply the Sum Rule:

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

= 2 (\int cos (u) d u - \int sin (3 u) d u)

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

\int sin (3 u) d u = - \frac{1}{3} cos (3 u)

= 2 (sin (u) - (- \frac{1}{3} cos (3 u)))

Substitute back

u = \frac{x}{2}

= 2 (sin (\frac{x}{2}) - (- \frac{1}{3} cos (3 \cdot \frac{x}{2})))

Simplify

2 (sin (\frac{x}{2}) - (- \frac{1}{3} cos (3 \cdot \frac{x}{2}))) : 2 (sin (\frac{x}{2}) + \frac{1}{3} cos (\frac{3 x}{2}))

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{7}{x ^{- 2}} + \frac{5}{x ^{- 3}})

x, \frac{1}{3} x, x = 25

\frac{d}{d x} (arccot (\frac{x}{5}))

x \to 0 lim (\frac{100}{2})

\frac{d}{d x} (tan^{4} (3) x^{2})

What is the integral of [cos(x/2)-sin(3 x/2)] ?
The integral of [cos(x/2)-sin(3 x/2)] is 2(sin(x/2)+1/3 cos((3x)/2))+C