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5.4. INTEGRATION BY PARTS
to
e t cos(t) dt. While the overall situation isn’t necessarily better than what we started
with, the problem hasn’t gotten worse. Thus, we proceed by integrating by parts again.
This time we let u = sin(t) and dv = e t dt, so that du = cos(t) dt and v = e t , which implies
e
t cos(t) dt = e
t cos(t) +
e
t sin(t) −
e
t cos(t) dt
(5.10)
We seem to be back where we started, as two applications of Integration by Parts has led
us back to the original problem,
e t cos(t) dt. But if we look closely at Equation (5.10),
we see that we can use algebra to solve for the value of the desired integral. In particular,
adding
e t cos(t) dt to both sides of the equation, we have
2
e
t cos(t) dt = e
t cos(t) + e
t sin(t),
and therefore
e
t cos(t) dt =
1
2
e
t cos(t) + e
t sin(t)
+ C.
Note that since we never actually encountered an integral we could evaluate directly, we
didn’t have the opportunity to add the integration constant C until the final step, at which
point we include it as part of the most general antiderivative that we sought from the
outset in evaluating an indefinite integral.
Activity 5.12.
Evaluate each of the following indefinite integrals.
(a)
x
2 sin(x) dx
(b)
t
3 ln(t) dt
(c)
e
z sin(z) dz
(d)
s
2 e
3s ds
(e)
t arctan(t) dt
(Hint: At a certain point in this problem, it is very helpful to note that
t 2
1+t 2 = 1 −
1
1+t 2 .)
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