What is the integral of 2tcos(3t) ?

The integral of 2tcos(3t) is 2/9 (3tsin(3t)+cos(3t))+C

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integral of 2tcos(3t)

\int 2 t cos (3 t) d t

\frac{2}{9} (3 t sin (3 t) + cos (3 t)) + C

Solution steps

\int 2 t cos (3 t) d t

Take the constant out:

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

= 2 \cdot \int t cos (3 t) d t

Apply u-substitution

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

= 2 \cdot \int \frac{u cos ( u )}{9} d u

Take the constant out:

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

= 2 \cdot \frac{1}{9} \cdot \int u cos (u) d u

Apply Integration By Parts

: u sin (u) - \int sin (u) d u

= 2 \cdot \frac{1}{9} (u sin (u) - \int sin (u) d u)

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

= 2 \cdot \frac{1}{9} (u sin (u) - (- cos (u)))

Substitute back

u = 3 t

= 2 \cdot \frac{1}{9} (3 t sin (3 t) - (- cos (3 t)))

Simplify

2 \cdot \frac{1}{9} (3 t sin (3 t) - (- cos (3 t))) : \frac{2}{9} (3 t sin (3 t) + cos (3 t))

= \frac{2}{9} (3 t sin (3 t) + cos (3 t))

Add a constant to the solution

= \frac{2}{9} (3 t sin (3 t) + cos (3 t)) + C

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What is the integral of 2tcos(3t) ?
The integral of 2tcos(3t) is 2/9 (3tsin(3t)+cos(3t))+C