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

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

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integral of-2x^2sin(pix)

\int - 2 x^{2} sin (π x) d x

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

Solution steps

\int - 2 x^{2} sin (π x) d x

Take the constant out:

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

= - 2 \cdot \int x^{2} sin (π x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

\int \frac{y}{y} d y

f (x) = - 1 - 2 cos (x), at x = - \frac{π}{4}

y^{^{'}} = (x - 3) (y + 1)^{\frac{2}{3}}, y (0) = - 1

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

(5 x - 4)^{4}, at x = 2

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