What is the integral of (-cos(3t)sin(3t))/3 ?

The integral of (-cos(3t)sin(3t))/3 is 1/18 cos^2(3t)+C

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integral of (-cos(3t)sin(3t))/3

\int \frac{- cos ( 3 t ) sin ( 3 t )}{3} d t

\frac{1}{18} cos^{2} (3 t) + C

Solution steps

\int \frac{- cos ( 3 t ) sin ( 3 t )}{3} d t

Take the constant out:

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

= \frac{- 1}{3} \cdot \int cos (3 t) sin (3 t) d t

Simplify

= (- \frac{1}{3}) \cdot \int cos (3 t) sin (3 t) d t

Apply u-substitution

: \int - \frac{u}{3} d u

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{u ^{2}}{2}

= (- \frac{1}{3}) (- \frac{1}{3} \cdot \frac{u ^{2}}{2})

Substitute back

u = cos (3 t)

= (- \frac{1}{3}) (- \frac{1}{3} \cdot \frac{cos ^{2} ( 3 t )}{2})

Simplify

(- \frac{1}{3}) (- \frac{1}{3} \cdot \frac{cos ^{2} ( 3 t )}{2}) : \frac{1}{18} cos^{2} (3 t)

= \frac{1}{18} cos^{2} (3 t)

Add a constant to the solution

= \frac{1}{18} cos^{2} (3 t) + C

\int \frac{x ln ( x ^{2} + 1 )}{x ^{2} + 1} d x

\int sec (a) x d x

\frac{\partial}{\partial x} (4 x^{3} + z^{2})

\frac{d}{d x} ((r (x) (x)^{2} - x^{2})^{\frac{1}{2}})

f (x) = sin (3 x), (\frac{π}{18}, \frac{2}{2})

What is the integral of (-cos(3t)sin(3t))/3 ?
The integral of (-cos(3t)sin(3t))/3 is 1/18 cos^2(3t)+C