What is the integral of-2ysin^2(y) ?

The integral of-2ysin^2(y) is -y^2+1/2 ysin(2y)+(y^2)/2+1/4 cos(2y)+C

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integral of-2ysin^2(y)

\int - 2 y sin^{2} (y) d y

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

Solution steps

\int - 2 y sin^{2} (y) d y

Take the constant out:

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

= - 2 \cdot \int y sin^{2} (y) d y

Apply Integration By Parts

: \frac{1}{2} y (y - \frac{1}{2} sin (2 y)) - \int \frac{1}{2} (y - \frac{1}{2} sin (2 y)) d y

= - 2 (\frac{1}{2} y (y - \frac{1}{2} sin (2 y)) - \int \frac{1}{2} (y - \frac{1}{2} sin (2 y)) d y)

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

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of-2ysin^2(y) ?
The integral of-2ysin^2(y) is -y^2+1/2 ysin(2y)+(y^2)/2+1/4 cos(2y)+C