What is the integral of 5cos(9t) ?

The integral of 5cos(9t) is 5/9 sin(9t)+C

Popular >

integral of 5cos(9t)

\int 5 cos (9 t) d t

\frac{5}{9} sin (9 t) + C

Solution steps

\int 5 cos (9 t) d t

Take the constant out:

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

= 5 \cdot \int cos (9 t) d t

Apply u-substitution

: \int cos (u) \frac{1}{9} d u

= 5 \cdot \int cos (u) \frac{1}{9} d u

Take the constant out:

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

= 5 \cdot \frac{1}{9} \cdot \int cos (u) d u

Use the common integral:

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

= 5 \cdot \frac{1}{9} sin (u)

Substitute back

u = 9 t

= 5 \cdot \frac{1}{9} sin (9 t)

Simplify

5 \cdot \frac{1}{9} sin (9 t) : \frac{5}{9} sin (9 t)

= \frac{5}{9} sin (9 t)

Add a constant to the solution

= \frac{5}{9} sin (9 t) + C

\int \frac{1}{x 4 x + 81} d x

\int \frac{( 5 x - 3 ) ^{3}}{x ^{3}} d x

\int - 2 π sin^{2} (2 x) d x

\int \frac{3 ( 1 + t ^{\frac{1}{2}} ) ^{2}}{2 t ^{\frac{1}{2}}} d t