What is the integral of-10cos(-2x+7) ?

The integral of-10cos(-2x+7) is 5sin(-2x+7)+C

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integral of-10cos(-2x+7)

\int - 10 cos (- 2 x + 7) d x

5 sin (- 2 x + 7) + C

Solution steps

\int - 10 cos (- 2 x + 7) d x

Take the constant out:

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

= - 10 \cdot \int cos (- 2 x + 7) d x

Apply u-substitution

: \int - \frac{1}{2} cos (u) d u

= - 10 \cdot \int - \frac{1}{2} cos (u) d u

Take the constant out:

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

= - 10 (- \frac{1}{2} \cdot \int cos (u) d u)

Use the common integral:

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

= - 10 (- \frac{1}{2} sin (u))

Substitute back

u = - 2 x + 7

= - 10 (- \frac{1}{2} sin (- 2 x + 7))

Simplify

- 10 (- \frac{1}{2} sin (- 2 x + 7)) : 5 sin (- 2 x + 7)

= 5 sin (- 2 x + 7)

Add a constant to the solution

= 5 sin (- 2 x + 7) + C

f (x) = sin (sin (x))

x^{2} d x = y^{3} d y

\frac{d}{d x} (lo g_{e} (3 x^{2} + 7))

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

\frac{d}{d x} (\frac{1 - 4 x}{3 + x})

What is the integral of-10cos(-2x+7) ?
The integral of-10cos(-2x+7) is 5sin(-2x+7)+C