What is the integral of cos(t-2x) ?

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

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integral of cos(t-2x)

\int cos (t - 2 x) d x

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

Solution steps

\int cos (t - 2 x) d x

Rewrite using trig identities

= \int cos (t) cos (2 x) + sin (t) sin (2 x) d x

Apply the Sum Rule:

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

= \int cos (t) cos (2 x) d x + \int sin (t) sin (2 x) d x

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

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

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

Use the Angle Difference identity:

sin (s) cos (t) - cos (s) sin (t) = sin (s - t)

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

Add a constant to the solution

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

\int (6 x^{2} - 3 x) d x

\frac{d}{d x} (lo g_{6} (2 x^{3} - 5 x^{2}))

\int tan^{2} (x se) c^{3} x d x

\frac{d}{d x} ((1 + x)^{15})

y^{^{''}} + 4 y = sin (2 t)

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