What is the integral from 1 to 4 of 3/2 sin(2-x/2) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 1 to 4 of 3/2 sin(2-x/2)

\int_{1}^{4} \frac{3}{2} sin (2 - \frac{x}{2}) d x

3 cos^{2} (2) - 3 cos (2) cos (\frac{1}{2}) + 3 sin^{2} (2) - 3 sin (2) sin (\frac{1}{2})

Decimal

2.78778 \dots

Solution steps

\int_{1}^{4} \frac{3}{2} sin (2 - \frac{x}{2}) d x

Take the constant out:

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

= \frac{3}{2} \cdot \int_{1}^{4} sin (2 - \frac{x}{2}) d x

Rewrite using trig identities

= \frac{3}{2} \cdot \int_{1}^{4} - cos (2) sin (\frac{x}{2}) + cos (\frac{x}{2}) sin (2) 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

= \frac{3}{2} (- \int_{1}^{4} cos (2) sin (\frac{x}{2}) d x + \int_{1}^{4} cos (\frac{x}{2}) sin (2) d x)

\int_{1}^{4} cos (2) sin (\frac{x}{2}) d x = - 2 cos^{2} (2) + 2 cos (2) cos (\frac{1}{2})

\int_{1}^{4} cos (\frac{x}{2}) sin (2) d x = 2 sin^{2} (2) - 2 sin (2) sin (\frac{1}{2})

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

Simplify

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

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

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