What is the integral of 5sin(2t) ?

The integral of 5sin(2t) is -5/2 cos(2t)+C

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integral of 5sin(2t)

\int 5 sin (2 t) d t

- \frac{5}{2} cos (2 t) + C

Solution steps

\int 5 sin (2 t) d t

Take the constant out:

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

= 5 \cdot \int sin (2 t) d t

Apply u-substitution

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

= 5 \cdot \int sin (u) \frac{1}{2} d u

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int sin (u) d u

Use the common integral:

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

= 5 \cdot \frac{1}{2} (- cos (u))

Substitute back

u = 2 t

= 5 \cdot \frac{1}{2} (- cos (2 t))

Simplify

5 \cdot \frac{1}{2} (- cos (2 t)) : - \frac{5}{2} cos (2 t)

= - \frac{5}{2} cos (2 t)

Add a constant to the solution

= - \frac{5}{2} cos (2 t) + C

f (3) = - \frac{2}{( x - 2 ) ^{2}}

\frac{d ^{2}}{d x ^{2}} (5 x^{2})

\frac{\partial}{\partial z} (sin (y) cos (x))

2 x - 17