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

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

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integral of cos(t-x)

\int cos (t - x) d x

- sin (t - x) + C

Solution steps

\int cos (t - x) d x

Apply u-substitution

: \int - cos (u) d u

= \int - cos (u) d u

Take the constant out:

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

= - \int cos (u) d u

Use the common integral:

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

= - sin (u)

Substitute back

u = t - x

= - sin (t - x)

Add a constant to the solution

= - sin (t - x) + C

f (x) = 7 x - 5

\int (2 x + 4)^{2} d x

\frac{d}{d x} (\frac{1}{ln ( 2 )})

y = \frac{7 x}{x - 1}, (2, 14)

\frac{\partial}{\partial x} ((x + 2)^{2} \cdot (1 + \frac{1}{y})^{3})