What is the integral of-9cos(3x+pi) ?

The integral of-9cos(3x+pi) is 3sin(3x)+C

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integral of-9cos(3x+pi)

\int - 9 cos (3 x + π) d x

3 sin (3 x) + C

Solution steps

\int - 9 cos (3 x + π) d x

Take the constant out:

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

= - 9 \cdot \int cos (3 x + π) d x

Rewrite using trig identities

= - 9 \cdot \int - cos (3 x) d x

Take the constant out:

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

= - 9 (- \int cos (3 x) d x)

Apply u-substitution

: \int cos (u) \frac{1}{3} d u

= - 9 (- \int cos (u) \frac{1}{3} d u)

Take the constant out:

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

= - 9 (- \frac{1}{3} \cdot \int cos (u) d u)

Use the common integral:

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

= - 9 (- \frac{1}{3} sin (u))

Substitute back

u = 3 x

= - 9 (- \frac{1}{3} sin (3 x))

Simplify

- 9 (- \frac{1}{3} sin (3 x)) : 3 sin (3 x)

= 3 sin (3 x)

Add a constant to the solution

= 3 sin (3 x) + C

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