What is the integral of (2x+8)sin(2x) ?

The integral of (2x+8)sin(2x) is 1/2 (-cos(2x)(2x+8)+sin(2x))+C

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integral of (2x+8)sin(2x)

\int (2 x + 8) sin (2 x) d x

\frac{1}{2} (- cos (2 x) (2 x + 8) + sin (2 x)) + C

Solution steps

\int (2 x + 8) sin (2 x) d x

Apply u-substitution

: \int (u + 8) sin (u) \frac{1}{2} d u

= \int (u + 8) sin (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int (u + 8) sin (u) d u

Apply Integration By Parts

: - cos (u) (u + 8) - \int - cos (u) d u

= \frac{1}{2} (- cos (u) (u + 8) - \int - cos (u) d u)

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

= \frac{1}{2} (- cos (u) (u + 8) - (- sin (u)))

Substitute back

u = 2 x

= \frac{1}{2} (- cos (2 x) (2 x + 8) - (- sin (2 x)))

Simplify

= \frac{1}{2} (- cos (2 x) (2 x + 8) + sin (2 x))

Add a constant to the solution

= \frac{1}{2} (- cos (2 x) (2 x + 8) + sin (2 x)) + C

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

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\int 4^{x + \frac{1}{2}} d x

\int - 2 x \cdot cos (x) d x

What is the integral of (2x+8)sin(2x) ?
The integral of (2x+8)sin(2x) is 1/2 (-cos(2x)(2x+8)+sin(2x))+C