What is the integral of-25cos(-5t) ?

The integral of-25cos(-5t) is -5sin(5t)+C

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integral of-25cos(-5t)

\int - 25 cos (- 5 t) d t

- 5 sin (5 t) + C

Solution steps

\int - 25 cos (- 5 t) d t

Take the constant out:

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

= - 25 \cdot \int cos (- 5 t) d t

Apply u-substitution

: \int - \frac{1}{5} cos (u) d u

= - 25 \cdot \int - \frac{1}{5} cos (u) d u

Take the constant out:

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

= - 25 (- \frac{1}{5} \cdot \int cos (u) d u)

Use the common integral:

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

= - 25 (- \frac{1}{5} sin (u))

Substitute back

u = - 5 t

= - 25 (- \frac{1}{5} sin (- 5 t))

Simplify

- 25 (- \frac{1}{5} sin (- 5 t)) : - 5 sin (5 t)

= - 5 sin (5 t)

Add a constant to the solution

= - 5 sin (5 t) + C

x \to - 1 lim (\frac{2}{x + 1})

2^{t^{3}}

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

\int \frac{1}{x 25 x ^{2} - 1} d x

x^{2} \frac{d y}{d x} - y x = 0