What is the integral of sin(3q)+cos(3q) ?

The integral of sin(3q)+cos(3q) is 1/3 (-cos(3q)+sin(3q))+C

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integral of sin(3q)+cos(3q)

\int sin (3 q) + cos (3 q) d q

\frac{1}{3} (- cos (3 q) + sin (3 q)) + C

Solution steps

\int sin (3 q) + cos (3 q) d q

Apply u-substitution

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

= \int (sin (u) + 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

= \frac{1}{3} \cdot \int sin (u) + cos (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int sin (u) d u + \int cos (u) d u)

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

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

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

Substitute back

u = 3 q

= \frac{1}{3} (- cos (3 q) + sin (3 q))

Add a constant to the solution

= \frac{1}{3} (- cos (3 q) + sin (3 q)) + C

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What is the integral of sin(3q)+cos(3q) ?
The integral of sin(3q)+cos(3q) is 1/3 (-cos(3q)+sin(3q))+C