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CALCULUS 1 PROBLEMS & SOLUTIONS

6.5.3
The Method Of Integration By Parts


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Review


1. Note

Let's find ò x cos x dx. We know that cos x is the derivative of sin x. So let's try the substitution u =
sin x, which yields du = cos x dx and x = arcsin u. Thus ò x cos x dx = ò arcsin u du. We don't
know yet how to find the integral of the function arcsin. Besides, the equation x = arcsin u requires
that x is in [–p/2, p/2], a restriction that's not given. Consequently, the method of substitution alone
doesn't seem to work with this integral.

So let's search for another way. We notice that the integrand is the product of two functions: x and
cos x. The product rule of differentiation is (uv)' = u'v + uv', where u and v are functions of x.
Hence, integrating both sides of this formula we get:

uv = ò u'v dx + ò uv' dx.

Note that we don't have to include a constant of integration on the left-hand side; this is because there
are still more than one indefinite integrals remaining in the equation. Each integrand on the right-hand
side is also a product of two functions. Putting ò x cos x dx to the place of ò u'v dx let u' = x and v =
cos x. It follows that u = (1/2)x² and v' = – sin x. Therefore, (1/2)x² cos x = ò x cos x dx –
(1/2) ò x² sin x dx, yielding ò x cos x dx = (1/2)x² cos x + (1/2) ò x² sin x dx. Well, we don't get
anywhere: the integral ò x² sin x dx appears to be harder than ò x cos x dx.

So let's switch and put ò x cos x dx to the place of ò uv' dx: let u = x and v' = cos x, yielding u' = 1
and v = sin x. Thus, x sin x = ò sin x dx + ò x cos x dx. Consequently:

ò x cos x dx = x sin x – ò sin x dx = x sin x + cos x + C.

Wow! Wait a minute, let's check to see if it's correct: (x sin x + cos x)' = sin x + x cos x – sin x =
x cos x. It's correct! Eureka!

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2. The Method Of Integration By Parts

The above successful solution is an example of the method of integration by parts. Let u = f(x) and
v = g(x) be two differentiable functions. By the product rule of differentiation we have (uv)' = u'v +
uv'. Integrating both sides of this formula we get uv = ò u'v dx + ò uv' dx. Note that we don't have to
include a constant of integration on the left-hand side; this is because there are still more than one
indefinite integrals remaining in the equation. We have u' dx = du and v' dx = dv. So uv = ò v du +
ò u dv. Consequently:




Example 1

Redo ò x cos x dx using Formula [1].

Solution
Let u = x and dv = cos x dx. So du = dx and v = sin x. It follows that:

ò x cos x dx = x sin x – ò sin x dx = x sin x + cos x + C.
EOS

Note that we don't write: ò x cos x dx = ò u dv = uv – ò v du = x sin x – ò sin x dx = x sin x +
cos x + C. Substituting u and v into the integral is not wrong, but not necessary either. So we skip it.
We simply use the formula ò u dv = uv – ò v du in our mind as a pattern or mnemonic device in
writing out the given integral according to that formula.

Let's see what happens if we instead let u = cos x and dv = x dx. Then du = – sin x dx and v =
(1/2)x². Thus, ò x cos x dx = (1/2)x² cos x + (1/2) ò x² sin x dx. The integral ò x² sin x dx is more
complicated than the original integral ò x cos x dx. So we stop here and try the choices u = x and
dv = cos x dx. And we find out that a new integral, ò sin x dx, is obtained that's simpler than the
original. Hence we continue our task, and, as expected, arrive at the answer.


General Case

Formula [1] can be used to find the integral of the product of two functions. Given the integral
ò f(x)g(x) dx to find, let's try the choices u = f(x) and dv = g(x) dx. Then du = f '(x) dx, and v’ =
g(x), so that v = ò g(x) dx. Obtaining du is a matter of differentiation, so there's no difficulty.
However, obtaining v is by integration, so it should readily be found. Using Formula [1] we get
ò f(x)g(x) dx = ò u dv = uv – ò v du. And we obtain a new integral to find: ò v du, which shouldn't
be more complicated than the original ò f(x)g(x) dx. When ò v du is found, our task is done.

If the choices u = f(x) and dv = g(x) dx lead to either v not being readily found or to ò v du being
more complicated than the original ò f(x)g(x) dx, quit and try the choices u = g(x) and dv = f(x) dx.

The technique of integration employing Formula [1] transforms a given integral, ò u dv, into the
difference between an already integrated term, uv, and another integral which will be integrated
afterward, ò v du. That is, this technique accomplishes the integration of one part of the given integral
at a time. Therefore, it's called the method of integration by parts, or the method of partial
integration.

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 2

Find ò x² cos x dx.

Solution
Let u = x² and dv = cos x dx. Then du = 2x dx and v = sin x. So:

ò x² cos x dx = x² sin x – ò sin x 2x dx = x² sin x – 2 ò x sin x dx.

Let u = x and dv = sin x dx. Thus du = dx and v = – cos x. Consequently:

ò x sin x dx = x (– cos x) – ò (– cos x) dx = – x cos x + ò cos x dx = – x cos x + sin x + C₁.

Hence:

ò x² cos x dx = x² sin x – 2(– x cos x + sin x) + C = x² sin x + 2 x cos x – 2 sin x + C.
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.

In the second application, if instead we let u = sin x and dv = x dx, then du = cos x dx and v =
(1/2)x². So ò x sin x dx = (1/2)x² sin x – (1/2) ò x² cos x dx. It follows that ò x² cos x dx = x² sin x –
x² sin x + ò x² cos x dx = ò x² cos x dx. Thus 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 as in the first.

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4. Regarding The Integrand As The Product Of Itself And 1

Example 3

Calculate ò ln x dx.

Solution
Let u = ln x and dv = dx, so that du = (1/x) dx and v = x. Thus:

ò ln x dx = (ln x) x – ò x (1/x) dx = x ln x – ò dx = x ln x – x + C.
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 4

Compute ò e^x sin x dx.

Solution
Let u = e^x and dv = sin x dx, so that du = e^x dx and v = – cos x. Thus:

ò e^x sin x dx = e^x (– cos x) – ò (– cos x) e^x dx = – e^x cos x + ò e^x cos x dx.

Let u = e^x and dv = cos x dx, so that du = e^x dx and v = sin x. Consequently:

ò e^x cos x dx = e^x sin x – ò (sin x) e^x dx = e^x sin x – ò e^x sin x dx.

Hence:

ò e^x sin x dx = – e^x cos x + e^x sin x – ò e^x sin x dx,

which leads to:

2 ò e^x sin x dx = – e^x cos x + e^x sin x + C₁.

Therefore:

ò e^x sin x dx = (1/2) (– e^x cos x + e^x sin x) + C = (1/2) e^x (sin x – cos x) + C.
EOS

The original integral ò e^x sin x dx has re-appeared, and we've obtained an equation where it's treated
as the unknown. We solve the equation, and the integral is found. The equation can be solved for the
original integral as long as the coefficient of the re-appearing term isn't 1 (if the coefficient is 1, the
re-appearing term and the term on the left-hand side will cancel out and both will disappear).

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6. Handling The Definite Integrals

Example 6



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

Derive this formula:

ò lnⁿ x dx = x lnⁿ x – n ò ln^n–1 x dx,

where n is a positive integer. Calculate ò ln⁴ x dx. Note that ln^k x = ( ln x)^k.



The above formula that gives ò lnⁿ x dx in terms of ò ln^n–1 x dx is called a reduction formula, because
it reduces the complexity of an integral. Note how the calculation of the integral with a particular value
of n (n = 4 in the above example) is carried out.

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8. Remarks

To apply the method of integration by parts to an integral, look for a factor that's readily or easily
integrated, and include it with dx to make up dv. The rest of the integrand is u. In using the formula
ò u dv = uv – ò v du, the resulting integral ò v du should be easier, or at least shouldn't be more
difficult, to integrate than the original integral ò u dv. An improper choice can make the original integral
more rather than less complicated and helpless. If things don't seem to work out, change the choices
for u and dv.

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).

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Problems & Solutions


1. Find the following integrals.



Solution

a. Let u = x and dv = sin 5x dx, so that du = dx and v = –(1/5) cos 5x. Thus:

b. Let u = x² and dv = x cos x² dx, so that du = 2x dx and v = (1/2) sin x². Consequently:

ò x³ cos x² dx = x² (1/2) sin x² – ò ((1/2) sin x²) 2x dx = (1/2) x² sin x² – ò x sin x² dx.

Let w = x², so that dw = 2x dx. Hence:

ò x sin x² dx = (1/2) ò sin w dw = – (1/2) cos w + C = – (1/2) cos x² + C.

It follows that:

ò x³ cos x² dx = (1/2) (x² sin x² + cos x²) + C.









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2. Find a reduction formula for ò xⁿe^ax dx.

Solution

Let u = xⁿ and dv = e^ax dx, so that du = nx^n–1 dx and v = (1/a) e^ax. Thus:

ò xⁿe^ax dx = xⁿ (1/a) e^ax – ò (1/a) e^ax nx^n–1 dx = (1/a) xⁿe^ax – (n/a) ò x^n–1e^ax dx.

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3. Let I_n = ò sinⁿ x dx, where n ³ 2.
a. Obtain a reduction formula for I_n.
b. Calculate I₅ and I₆.

Solution



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Solution





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