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

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

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integral of sin^2(x-y)

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

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

Solution steps

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

Rewrite using trig identities

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

Take the constant out:

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

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

x - y = x (1 - \frac{y}{x})

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

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

\int 1 d y = y

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

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

Simplify

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

Add a constant to the solution

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

y = x^{2} + 2 x + 10, at x = 6

\frac{d y}{d x} = 2 y + x^{2} e^{2 x} + 5

x \to 0 lim (2400 x - 2 x^{2})

x \to 1 lim (\frac{x}{( x - 1 )})

y = 6 x^{3} - x^{2}

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