What is the integral of sin(t)cos(t-x) ?

The integral of sin(t)cos(t-x) is -sin(t)sin(t-x)+C

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integral of sin(t)cos(t-x)

\int sin (t) cos (t - x) d x

- sin (t) sin (t - x) + C

Solution steps

\int sin (t) cos (t - x) d x

Take the constant out:

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

= sin (t) \cdot \int cos (t - x) d x

Apply u-substitution

: \int - cos (u) d u

= sin (t) \cdot \int - cos (u) d u

Take the constant out:

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

= sin (t) (- \int cos (u) d u)

Use the common integral:

\int cos (u) d u = sin (u)

= sin (t) (- sin (u))

Substitute back

u = t - x

= sin (t) (- sin (t - x))

Simplify

= - sin (t) sin (t - x)

Add a constant to the solution

= - sin (t) sin (t - x) + C

\int x 10 - x^{2} d x

\frac{d}{d x} (\frac{1}{6} x e^{3 x})

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

\int \frac{6 x ^{3}}{16 + x ^{2}} d x

y^{^{'}} = 2 y + x^{2} + 9

What is the integral of sin(t)cos(t-x) ?
The integral of sin(t)cos(t-x) is -sin(t)sin(t-x)+C