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Calculus 1 Problems & Solutions – Chapter 6 – Section 6.5.3 |
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6.5.3
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Review |
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1. Introduction |
Wow! … Wait a minute,
let's check to see if it's correct: (x sin x + cos x + C)' = (1)sin x + x cos
x – sin x + 0 =
x cos x. It's
correct!
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2. The Method Of Integration By Parts |
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Example 2.1
Calculate:
Solution
Let u = x and dv = cos x dx. Then du = dx and v = sin x. So:
EOS
General Case
Formula [2.1] can be used to find the integral of the product of 2 functions. Suppose that we're given the integral:
The constant of integration for v disappears. As a consequence
it isn't necessary. It can only be cumbersome, and hence
should be omitted.
Recall that the method of substitution, a feature of
integration, is derived from and inverse to the chain rule, a feature of
differentiation. Now we see that the method of integration by parts, a feature
of integration, is derived from and inverse
to the product rule, a feature of differentiation (differentiating uv to get to u'v + uv'; integrating u'v + uv' to get back to
uv).
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3. Applying More Than Once |
Example 3.1
Compute:
Solution
where C
= –2C1.
EOS
Here we apply the method of integration by parts twice.
First, we apply it to the original integral, and second, to the
integral obtained from the first application. We use the same letters u and v in both
steps. Don't be confused. It's ok to do
so. Think of this situation as assigning new values to u and v.
We would have undone what we've done in the first step. This
shows that we should make a similar choice for u in the
second application, u
= x,
as in the first, u
= x2.
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4. Regarding The Integrand As The Product Of Itself And 1 |
Example 4.1
Evaluate:
Solution
Let u
= ln x and dv = dx, so that du = (1/x) dx and v
= x. Thus:
EOS
We have to regard the integrand as the product of itself and
1: ln x = (ln x) . 1. So we
choose u
= ln x and dv = 1 . dx
= dx.
Looking at it in another way, we can say that sometimes it's necessary to
choose dv
= dx
only.
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5. Re-Appearance Of The Original Integral |
Example 5.1
Find:
Solution
Let u
= ex and dv = sin x dx, so that du = ex dx and v = – cos x. Thus:
where C = (1/2)C1.
EOS
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6. Handling The Definite Integrals |
Example 6.1
Calculate:
Solution
EOS
When handling definite integrals, remember to include the evaluation symbol with any term that's been integrated.
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7. Reduction Formulas |
Example 7.1
Derive this formula:
Solution
Let u
= lnn x
and dv
= dx,
so that du
= (n/x) lnn–1 x dx and v = x. Thus:
EOS
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8. Remarks |
The preceding examples show that if the integrand involves a
power of x
or an inverse trigonometric function or a
logarithm (which is the inverse of the exponential function), then we should
let u
be the power of x
(to lower the power)
or the inverse trigonometric function or the logarithm (to transform the
integrand into an algebraic expression). See
Example 3.1 for a power of x, Example
6.1 for an inverse trigonometric function, and Example
4.1 for a logarithm.
Problems & Solutions |
1. Calculate the following integrals.
Solution
a. Let u = x and dv = sin 5x dx, so that du = dx and v = –(1/5) cos 5x. Thus:
It follows that:
where C = (9/13)C1.
2. Compute this definite integral:
Solution
It follows that:
3. Find a reduction formula for:
Solution
Let u = xn and dv = eax dx, so that du = nxn–1 dx and v = (1/a) eax. Thus:
4. Let:
Solution
a. Let u = sinn–1 x and dv = sin x dx, so that du = (n – 1) sinn–2 x cos x dx and v = – cos x. Then:
where C = (3/4)C1,
where C = (5/6)C1.
5. Let:
Solution
c. We have:
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