What is the integral of 2t*(-cos(2t)) ?

The integral of 2t*(-cos(2t)) is -tsin(2t)-1/2 cos(2t)+C

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integral of 2t*(-cos(2t))

\int 2 t \cdot (- cos (2 t)) d t

- t sin (2 t) - \frac{1}{2} cos (2 t) + C

Solution steps

\int 2 t (- cos (2 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 (2 t)) d t

Rewrite using trig identities

= 2 \cdot \int - t (cos^{2} (t) - sin^{2} (t)) d t

Take the constant out:

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

= 2 (- \int t (cos^{2} (t) - sin^{2} (t)) d t)

Rewrite using trig identities

= 2 (- \int t cos (2 t) d t)

Apply u-substitution

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

= 2 (- \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

= 2 (- \frac{1}{4} \cdot \int u cos (u) d u)

Apply Integration By Parts

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

= 2 (- \frac{1}{4} (u sin (u) - \int sin (u) d u))

Use the common integral:

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

= 2 (- \frac{1}{4} (u sin (u) - (- cos (u))))

Substitute back

u = 2 t

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

Simplify

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

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

Add a constant to the solution

= - t sin (2 t) - \frac{1}{2} cos (2 t) + C

f (x) = e^{x}, at x = 5

x \to 0 lim (\frac{2}{1 - 2 ^{\frac{1}{x}}})

t^{3} \frac{d y}{d t} + 4 t^{2} y = e^{- t}, y (- 1) = 0

f (x) = \frac{2}{x - 1}

- \frac{5}{s ( 50 + 20 s )}

What is the integral of 2t*(-cos(2t)) ?
The integral of 2t*(-cos(2t)) is -tsin(2t)-1/2 cos(2t)+C