Hi

I have to find the indefinite integral of:

f(t) = 6cos(3t)+5e^-10t

can anyone please show me step by step how to work this out.

many thanks

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- Mar 3rd 2008, 11:57 AMtracy207Indefinite integrals
Hi

I have to find the indefinite integral of:

f(t) = 6cos(3t)+5e^-10t

can anyone please show me step by step how to work this out.

many thanks - Mar 3rd 2008, 01:07 PMtopsquark
$\displaystyle \int f(t)~dt = 6 \int cos(3t)~dt + \int e^{-10t}~dt$

The first integral may be found by substituting $\displaystyle u = 3t \implies du = 3 dt$:

$\displaystyle \int cos(3t)~dt = \int cos(u) \cdot \frac{du}{3} = \frac{1}{3} \int cos(u)~du = \frac{1}{3} \cdot sin(u) = \frac{1}{3} \cdot sin(3t)$

For the second integral, use $\displaystyle u = -10t \implies du = -10 dt$.

You give this part a try. (And, of course, don't leave out that arbitrary constant at the end.)

-Dan