What is the integral of 8sin(-2x+7) ?

The integral of 8sin(-2x+7) is 4cos(-2x+7)+C

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integral of 8sin(-2x+7)

\int 8 sin (- 2 x + 7) d x

4 cos (- 2 x + 7) + C

Solution steps

\int 8 sin (- 2 x + 7) d x

Take the constant out:

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

= 8 \cdot \int sin (- 2 x + 7) d x

Apply u-substitution

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

= 8 \cdot \int - \frac{1}{2} sin (u) d u

Take the constant out:

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

= 8 (- \frac{1}{2} \cdot \int sin (u) d u)

Use the common integral:

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

= 8 (- \frac{1}{2} (- cos (u)))

Substitute back

u = - 2 x + 7

= 8 (- \frac{1}{2} (- cos (- 2 x + 7)))

Simplify

8 (- \frac{1}{2} (- cos (- 2 x + 7))) : 4 cos (- 2 x + 7)

= 4 cos (- 2 x + 7)

Add a constant to the solution

= 4 cos (- 2 x + 7) + C

x \to 3 + lim (x - 3)

x \to 1 lim (2 x - 1)

y = x^{3}, y = x^{2} + x

\int 4 x + \frac{1}{x} d x

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

What is the integral of 8sin(-2x+7) ?
The integral of 8sin(-2x+7) is 4cos(-2x+7)+C