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

The integral of-tcos(2t) is -1/4 (2tsin(2t)+cos(2t))+C

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integral of-tcos(2t)

\int - t cos (2 t) d t

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

Solution steps

\int - t cos (2 t) d t

Take the constant out:

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

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

Apply u-substitution

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

= - \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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = 2 t

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

Simplify

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

Add a constant to the solution

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

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What is the integral of-tcos(2t) ?
The integral of-tcos(2t) is -1/4 (2tsin(2t)+cos(2t))+C