integral from -pi to pi of x*sin(x/2)

1) Let's integrate:

\int_{- π}^{π} x \cdot \sin (\frac{x}{2}) d x

⇥For

x \cdot \sin (\frac{x}{2})

, we use integration by parts:

\int u1 \cdot v2' d x = u1 \cdot v2 - \int u1' \cdot v2 d x

, with u1 =

x

and v2' =

\sin (\frac{x}{2})

⇥Thus,

u1'

1

⇥1.1) Let's integrate v2':

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

⇥For the math expression:

\sin (\frac{x}{2})

, since (cos(a))' = -a'*sin(a), we try differentiating

\cos (\frac{x}{2})

⇥1.2) Let's perform the differentiation:

\frac{d}{d x} (\cos (\frac{x}{2}))

⇥Apply the differentiation rule:

\frac{d}{d x} (\cos (f)) = - \frac{d}{d x} (f) \cdot \sin (f)

with f =

\frac{x}{2}

⇥ⓘ The derivative of cos(x) is -sin(x):

\frac{d}{d x} (\cos (\frac{x}{2}))

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

⇥

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

⇥ⓘ A constant factor stays in front of the derivative, (k*u)' = k*u'. The derivative of x is 1:

\frac{d}{d x} (\frac{x}{2})

\frac{1}{2}

⇥

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

⇥ⓘ Multiplying a fraction by a term multiplies its numerator:

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

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

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

⇥

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

⇥You observe that the computed derivative and the original math expression are the same, except for a coefficient. So the integral is:

⇥ⓘ The derivative of

-2 \cdot \cos (\frac{x}{2})

\sin (\frac{x}{2})

, so

-2 \cdot \cos (\frac{x}{2})

is an antiderivative of

\sin (\frac{x}{2})

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

-2 \cdot \cos (\frac{x}{2})

⇥

-2 \cdot \cos (\frac{x}{2})

⇥So,

v2

corresponds to

- 2 \cdot \cos (\frac{x}{2})

⇥1.3) Let's calculate

u1 \cdot v2 - \int u1' \cdot v2 d x

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

⇥For the math expression:

2 \cdot \cos (\frac{x}{2})

, since (sin(a))' = a'*cos(a), we try differentiating

\sin (\frac{x}{2})

⇥1.4) Let's differentiate:

\frac{d}{d x} (\sin (\frac{x}{2}))

⇥Apply the differentiation rule:

\frac{d}{d x} (\sin (f)) = \frac{d}{d x} (f) \cdot \cos (f)

with f =

\frac{x}{2}

⇥ⓘ The derivative of sin(x) is cos(x):

\frac{d}{d x} (\sin (\frac{x}{2}))

\frac{d}{d x} (\frac{x}{2}) \cdot \cos (\frac{x}{2})

⇥

\frac{d}{d x} (\frac{x}{2}) \cdot \cos (\frac{x}{2})

⇥ⓘ A constant factor stays in front of the derivative, (k*u)' = k*u'. The derivative of x is 1:

\frac{d}{d x} (\frac{x}{2})

\frac{1}{2}

⇥

\frac{1}{2} \cdot \cos (\frac{x}{2})

⇥ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{1}{2} \cdot \cos (\frac{x}{2})

\frac{1 \cdot \cos (\frac{x}{2})}{2}

\frac{\cos (\frac{x}{2})}{2}

⇥

\frac{\cos (\frac{x}{2})}{2}

⇥You observe that the computed derivative and the original math expression are the same, except for a coefficient. So the integral is:

⇥ⓘ The derivative of

4 \cdot \sin (\frac{x}{2})

2 \cdot \cos (\frac{x}{2})

, so

4 \cdot \sin (\frac{x}{2})

is an antiderivative of

2 \cdot \cos (\frac{x}{2})

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

4 \cdot \sin (\frac{x}{2})

⇥

- 2 x \cdot \cos (\frac{x}{2}) + 4 \cdot \sin (\frac{x}{2})

Then you can resume the integration:

ⅰ

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

- 2 x \cdot \cos (\frac{x}{2}) + 4 \cdot \sin (\frac{x}{2})

▷

- 2 x \cdot \cos (\frac{x}{2}) + 4 \cdot \sin (\frac{x}{2})

2) Let's evaluate for the interval

[- π, π]

(4 \cdot \sin (\frac{π}{2}) - 2 π \cdot \cos (\frac{π}{2})) - (4 \cdot \sin (- \frac{π}{2}) - 2 \cdot (- π) \cdot \cos (- \frac{π}{2}))

ⓘ The sine is the y-coordinate (vertical axis) of the point reached on the unit circle after a rotation of

\frac{π}{2}

radians from the horizontal axis pointing to the right, this is a remarkable value:

\sin (\frac{π}{2})

1

■Multiplying by 1 changes nothing:

4 \cdot 1

4

■The cosine is the x-coordinate (horizontal axis) of the point reached on the unit circle after a rotation of

\frac{π}{2}

radians from the horizontal axis pointing to the right, this is a remarkable value:

\cos (\frac{π}{2})

0

■A minus sign before a parenthesis changes the sign of each of its terms:

- (4 \cdot \sin (- \frac{π}{2}) - 2 \cdot (- π) \cdot \cos (- \frac{π}{2}))

- 4 \cdot \sin (- \frac{π}{2}) + 2 \cdot (- π) \cdot \cos (- \frac{π}{2})

4

- 2 π \cdot 0

- 4 \cdot \sin (- \frac{π}{2})

+ 2 \cdot (- π) \cdot \cos (- \frac{π}{2})

ⓘ A product with a zero factor is zero:

2 π \cdot 0

0

■sin is odd, sin(-x) = -sin(x):

\sin (- \frac{π}{2})

- \sin (\frac{π}{2})

■cos is even, cos(-x) = cos(x):

\cos (- \frac{π}{2})

\cos (\frac{π}{2})

■The minus sign is placed in front of the product:

2 \cdot (- π)

-2 π

4

+ 0

- 4 \cdot (- \sin (\frac{π}{2}))

- 2 π \cdot \cos (\frac{π}{2})

ⓘ Adding 0 changes nothing:

0 - 4 \cdot (- \sin (\frac{π}{2}))

-4 \cdot (- \sin (\frac{π}{2}))

■The sine is the y-coordinate (vertical axis) of the point reached on the unit circle after a rotation of

\frac{π}{2}

radians from the horizontal axis pointing to the right, this is a remarkable value:

- \sin (\frac{π}{2})

-1

■Multiplying by -1 gives the opposite:

-4 \cdot (-1)

4

■The cosine is the x-coordinate (horizontal axis) of the point reached on the unit circle after a rotation of

\frac{π}{2}

radians from the horizontal axis pointing to the right, this is a remarkable value:

\cos (\frac{π}{2})

0

4 + 4 - 2 π \cdot 0

ⓘ Start from

4

and move forward

4

on the number line:

4 + 4

8

■A product with a zero factor is zero:

2 π \cdot 0

0

8 + 0

ⓘ Adding 0 changes nothing:

8 + 0

8

▶Integral:

8