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

The integral of cos(x)sin(t-x) is cos(t-x)cos(x)+1/4 (sin(t)(2x-t)-cos(2x-t))+C

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

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

cos (t - x) cos (x) + \frac{1}{4} (sin (t) (2 x - t) - cos (2 x - t)) + C

Solution steps

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

Apply Integration By Parts

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

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

\int - cos (t - x) sin (x) d x = - \frac{1}{4} (sin (t) (2 x - t) - cos (2 x - t))

= cos (t - x) cos (x) - (- \frac{1}{4} (sin (t) (2 x - t) - cos (2 x - t)))

Simplify

= cos (t - x) cos (x) + \frac{1}{4} (sin (t) (2 x - t) - cos (2 x - t))

Add a constant to the solution

= cos (t - x) cos (x) + \frac{1}{4} (sin (t) (2 x - t) - cos (2 x - t)) + C

\frac{\partial}{\partial x} (\frac{10 - x ^{3} - y ^{2}}{12 x})

\int_{- π}^{π} sin (2 x) sin (2 x) d x

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

\int x^{2} 5 x + 5 d x

\int \frac{5}{33 x} x^{4} d x

What is the integral of cos(x)sin(t-x) ?
The integral of cos(x)sin(t-x) is cos(t-x)cos(x)+1/4 (sin(t)(2x-t)-cos(2x-t))+C