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

The integral of sin(x)(-8xcos(2x)) is -8(1/2 x(-1/3 cos(3x)+cos(x))-1/2 (-1/9 sin(3x)+sin(x)))+C

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integral of sin(x)(-8xcos(2x))

\int sin (x) (- 8 x cos (2 x)) d x

- 8 (\frac{1}{2} x (- \frac{1}{3} cos (3 x) + cos (x)) - \frac{1}{2} (- \frac{1}{9} sin (3 x) + sin (x))) + C

Solution steps

\int sin (x) (- 8 x cos (2 x)) d x

Rewrite using trig identities

= \int - 8 x sin (x) (cos^{2} (x) - sin^{2} (x)) d x

Take the constant out:

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

= - 8 \cdot \int x sin (x) (cos^{2} (x) - sin^{2} (x)) d x

Use the Double Angle identity:

cos^{2} (x) - sin^{2} (x) = cos (2 x)

= - 8 \cdot \int x sin (x) cos (2 x) d x

Apply Integration By Parts

: \frac{1}{2} x (- \frac{1}{3} cos (3 x) + cos (x)) - \int \frac{1}{2} (- \frac{1}{3} cos (3 x) + cos (x)) d x

= - 8 (\frac{1}{2} x (- \frac{1}{3} cos (3 x) + cos (x)) - \int \frac{1}{2} (- \frac{1}{3} cos (3 x) + cos (x)) d x)

\int \frac{1}{2} (- \frac{1}{3} cos (3 x) + cos (x)) d x = \frac{1}{2} (- \frac{1}{9} sin (3 x) + sin (x))

= - 8 (\frac{1}{2} x (- \frac{1}{3} cos (3 x) + cos (x)) - \frac{1}{2} (- \frac{1}{9} sin (3 x) + sin (x)))

Add a constant to the solution

= - 8 (\frac{1}{2} x (- \frac{1}{3} cos (3 x) + cos (x)) - \frac{1}{2} (- \frac{1}{9} sin (3 x) + sin (x))) + C

x \to 0 lim (3 x + 2)

f (x) = \frac{2}{3 x}

2 x e^{- 2 x}

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

\int sin (8 x + 2) d x

What is the integral of sin(x)(-8xcos(2x)) ?
The integral of sin(x)(-8xcos(2x)) is -8(1/2 x(-1/3 cos(3x)+cos(x))-1/2 (-1/9 sin(3x)+sin(x)))+C