What is the integral of-tcos(t) ?

The integral of-tcos(t) is -tsin(t)-cos(t)+C

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integral of-tcos(t)

\int - t cos (t) d t

- t sin (t) - cos (t) + C

Solution steps

\int - t cos (t) d t

Take the constant out:

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

= - \int t cos (t) d t

Apply Integration By Parts

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

= - (t sin (t) - \int sin (t) d t)

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

= - (t sin (t) - (- cos (t)))

Simplify

= - t sin (t) - cos (t)

Add a constant to the solution

= - t sin (t) - cos (t) + C

f (x) = \frac{1}{x ^{2}}, (1, 1)

\frac{d}{d x} (\frac{10}{x ^{2} - 4 x + 1})

\int (x + 7) (x - 3) d x

\int 4 x - 4 d x

\frac{d}{d x} (\frac{e ^{x^{2}}}{x ^{3}})