What is the integral of (x^2-1)sin(2x) ?

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

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integral of (x^2-1)sin(2x)

\int (x^{2} - 1) sin (2 x) d x

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

Solution steps

\int (x^{2} - 1) sin (2 x) d x

Apply Integration By Parts

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

= - \frac{1}{2} cos (2 x) (x^{2} - 1) - \int - x cos (2 x) d x

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d m} (im)

y = \frac{15}{ln ( x )}

\frac{d}{d x} (8 e^{\frac{x}{2}})

4^{- x}

\frac{d ^{2}}{d x ^{2}} (ln (1 + x^{2}))

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