Calculus Of One Real Variable –
By Pheng Kim Ving 
10.8 
1. Introduction 
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yielding:
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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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Calculate:
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Formula [2.1] can be used to find the integral of the product of 2 functions. Suppose that we're given the integral:
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The constant of integration for v disappears. As a consequence it isn't
necessary. It can only be cumbersome, and hence
should be omitted.
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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 
Compute:
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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.
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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 = x^{2}.
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4. Regarding The Integrand As The Product Of Itself And 1 
Evaluate:
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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. ReAppearance Of The Original Integral 
Find:
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Solution
Let u = e^{x} and
dv = sin x
dx, so that du = e^{x} dx and v
= – cos x.
Thus:
where C = (1/2)C_{1}.
EOS
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6. Handling The Definite Integrals 
Calculate:
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When handling definite integrals, remember to include the evaluation symbol with any term that's been integrated.
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7. Reduction Formulas 
Derive this formula:
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Solution
Let u = ln^{n} x
and dv = dx, so that du = (n/x)ln^{n}^{–1} x dx and v
= x. Thus:
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EOS
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8. Remarks 
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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.
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Solution
a. Let u = x and dv = sin 5x dx, so that du = dx and v = –(1/5) cos 5x. Thus:
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where C = – C_{1}.
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It follows that:
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Let u = e^{2}^{x} and dv = cos 3x dx, so that du = 2e^{2}^{x} dx and v = (1/3) sin 3x. Thus:
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where C = (9/13)C_{1}.
2. Compute this definite integral:
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Solution
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}
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3. Find a reduction formula for:
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Solution
Let u = x^{n} and dv = e^{ax}^{ }dx, so that du = nx^{n}^{–1} dx and v = (1/a) e^{ax}. Thus:
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4. Let:
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Solution
a. Let u = sin^{n}^{–1} x and dv = sin x dx, so that du = (n – 1)sin^{n}^{–2} x cos x dx and v = – cos x. Then:
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where C = (3/4)C_{1},
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where C = (5/6)C_{1}.
5. Let:
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Solution
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c. We have:
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