What is the integral of sin(8-x) ?

The integral of sin(8-x) is cos(8-x)+C

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integral of sin(8-x)

\int sin (8 - x) d x

cos (8 - x) + C

Solution steps

\int sin (8 - x) d x

Apply u-substitution

: \int - sin (u) d u

= \int - sin (u) d u

Take the constant out:

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

= - \int sin (u) d u

Use the common integral:

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

= - (- cos (u))

Substitute back

u = 8 - x

= - (- cos (8 - x))

Simplify

= cos (8 - x)

Add a constant to the solution

= cos (8 - x) + C

\int_{- 1}^{0} x^{2} d x

\int e^{c o s (8 x)} sin (8 x) d x

x \to 1 lim ((x - 1) (x + 1))

\frac{d}{d x} (\frac{1}{cos ( x )})

x \to 0 lim (x e^{x})