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

1) Let's integrate:

\int_{- π}^{π} x \cdot \sin (2 x) d x

⇥For

x \cdot \sin (2 x)

, 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 (2 x)

⇥So the value of

u1'

1

⇥1.1) Let's compute the integral v2':

\int \sin (2 x) d x

⇥To use the known integral of

\sin (u)

, let

u = 2 x

. Then

du = 2 dx

, so

dx = \frac{1}{2} \cdot du

⇥We use

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

, then substitute

u = 2 x

back

⇥ⓘ The derivative of

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

\sin (2 x)

, so

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

is an antiderivative of

\sin (2 x)

\int \sin (2 x) d x

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

⇥

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

⇥So the value of

v2

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

⇥1.2) Let's calculate

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

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

⇥For the math expression:

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

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

\sin (2 x)

⇥1.3) Let's differentiate:

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

⇥Apply the differentiation rule:

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

with f =

2 x

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

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

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

⇥

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

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

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

2

⇥

2 \cdot \cos (2 x)

⇥The computed derivative matches the original math expression up to a constant factor. Therefore, the integral is:

⇥ⓘ The derivative of

\frac{1}{4} \cdot \sin (2 x)

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

, so

\frac{1}{4} \cdot \sin (2 x)

is an antiderivative of

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

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

\frac{1}{4} \cdot \sin (2 x)

⇥

- \frac{x \cdot \cos (2 x)}{2} + \frac{1}{4} \cdot \sin (2 x)

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

\frac{1}{4} \cdot \sin (2 x)

\frac{1 \cdot \sin (2 x)}{4}

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

■To add fractions, put them over the same denominator, then add the numerators:

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

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

⇥

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

⇥ⓘ The minus sign is placed in front of the product:

2 \cdot (- x)

-2 x

⇥

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

Then you can continue the integral computation:

ⅰ

\int x \cdot \sin (2 x) d x

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

▷

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

2) Let's evaluate for the interval

[- π, π]

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

ⓘ cos and sin are periodic with period 2π:

\sin (2 π)

\sin (0)

\cos (2 π)

\cos (0)

\sin (2 \cdot (- π))

\sin (0)

\cos (2 \cdot (- π))

\cos (0)

■Two minus signs cancel out:

-2 \cdot (- π)

2 π

\frac{\sin (0) - 2 π \cdot \cos (0)}{4} - \frac{\sin (0) + 2 π \cdot \cos (0)}{4}

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

0

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

\sin (0)

0

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

0

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

\cos (0)

1

■Multiplying by 1 changes nothing:

2 π \cdot 1

2 π

\frac{0 - 2 π}{4} - \frac{0 + 2 π}{4}

ⓘ Adding 0 changes nothing:

0 - 2 π

- 2 π

0 + 2 π

2 π

■Like terms are grouped by adding their coefficients:

- \frac{2 π}{4} - \frac{2 π}{4}

- π

▶Integral:

- π

3) The numeric value of:

- π

equals:

▷

\approx -3.1416