What is the integral of 2sin^2(y) ?

The integral of 2sin^2(y) is y-1/2 sin(2y)+C

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integral of 2sin^2(y)

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

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

Solution steps

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

Rewrite using trig identities

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

Take the constant out:

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d y = y

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} = \frac{y ^{2} - 1}{x ^{2} - 1}, y (2) = - 1

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

\frac{\partial}{\partial y} (1 + ln (x y))

\int_{- 1}^{1} \frac{3}{x ^{2}} d x

\frac{\partial}{\partial x} (L + (L^{2} + x^{2})^{\frac{1}{2}})

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