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

The integral of 3tcos(2t) is 3/4 (2tsin(2t)+cos(2t))+C

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integral of 3tcos(2t)

\int 3 t cos (2 t) d t

\frac{3}{4} (2 t sin (2 t) + cos (2 t)) + C

Solution steps

\int 3 t cos (2 t) d t

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u cos ( u )}{4} d u

= 3 \cdot \int \frac{u cos ( u )}{4} d u

Take the constant out:

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

= 3 \cdot \frac{1}{4} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= 3 \cdot \frac{1}{4} (u sin (u) - \int sin (u) d u)

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

= 3 \cdot \frac{1}{4} (u sin (u) - (- cos (u)))

Substitute back

u = 2 t

= 3 \cdot \frac{1}{4} (2 t sin (2 t) - (- cos (2 t)))

Simplify

3 \cdot \frac{1}{4} (2 t sin (2 t) - (- cos (2 t))) : \frac{3}{4} (2 t sin (2 t) + cos (2 t))

= \frac{3}{4} (2 t sin (2 t) + cos (2 t))

Add a constant to the solution

= \frac{3}{4} (2 t sin (2 t) + cos (2 t)) + C

x \to 8 lim (\frac{∣ 8 - x ∣}{8 - x})

\int \frac{x ^{2} + 1}{x ^{2} + 2} d x

x \to 3 - lim (\frac{7 - x}{x - 3})

\int \frac{x}{x ^{4} - 1} d x

\frac{\partial}{\partial x} (x^{2} y - y^{2} + y^{3} x^{2})

What is the integral of 3tcos(2t) ?
The integral of 3tcos(2t) is 3/4 (2tsin(2t)+cos(2t))+C