What is the integral of tcos(8t) ?

The integral of tcos(8t) is 1/64 (8tsin(8t)+cos(8t))+C

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integral of tcos(8t)

\int t cos (8 t) d t

\frac{1}{64} (8 t sin (8 t) + cos (8 t)) + C

Solution steps

\int t cos (8 t) d t

Apply u-substitution

: \int \frac{u cos ( u )}{64} d u

= \int \frac{u cos ( u )}{64} d u

Take the constant out:

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

= \frac{1}{64} \cdot \int u cos (u) d u

Apply Integration By Parts

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

= \frac{1}{64} (u sin (u) - \int sin (u) d u)

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

= \frac{1}{64} (u sin (u) - (- cos (u)))

Substitute back

u = 8 t

= \frac{1}{64} (8 t sin (8 t) - (- cos (8 t)))

Simplify

= \frac{1}{64} (8 t sin (8 t) + cos (8 t))

Add a constant to the solution

= \frac{1}{64} (8 t sin (8 t) + cos (8 t)) + C

\int - y + 1 d y

\frac{d}{d x} ((4 x^{3} + 5 x) (x - 3) (x + 5))

\frac{d}{d x} (5 arcsin (3) x^{3})

\frac{1}{11 x}

y = (1 + \frac{1}{x})^{2}

What is the integral of tcos(8t) ?
The integral of tcos(8t) is 1/64 (8tsin(8t)+cos(8t))+C