What is the integral of-40sin(5y+1) ?

The integral of-40sin(5y+1) is 8cos(5y+1)+C

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integral of-40sin(5y+1)

\int - 40 sin (5 y + 1) d y

8 cos (5 y + 1) + C

Solution steps

\int - 40 sin (5 y + 1) d y

Take the constant out:

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

= - 40 \cdot \int sin (5 y + 1) d y

Apply u-substitution

= - 40 \cdot \int sin (u) \frac{1}{5} d u

Take the constant out:

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

= - 40 \cdot \frac{1}{5} \cdot \int sin (u) d u

Use the common integral:

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

= - 40 \cdot \frac{1}{5} (- cos (u))

Substitute back

u = 5 y + 1

= - 40 \cdot \frac{1}{5} (- cos (5 y + 1))

Simplify

- 40 \cdot \frac{1}{5} (- cos (5 y + 1)) : 8 cos (5 y + 1)

= 8 cos (5 y + 1)

Add a constant to the solution

= 8 cos (5 y + 1) + C

What is the integral of-40sin(5y+1) ?
The integral of-40sin(5y+1) is 8cos(5y+1)+C