7 techniques of integration. due to the fundamental theorem of calculus (ftc), we can integrate a...
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7TECHNIQUES OF INTEGRATIONTECHNIQUES OF INTEGRATION
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TECHNIQUES OF INTEGRATION
Due to the Fundamental Theorem of Calculus
(FTC), we can integrate a function if we know
an antiderivative, that is, an indefinite integral.
We summarize the most important integrals we have learned so far, as follows.
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FORMULAS OF INTEGRALS
1 1( 1) ln | |
1
ln
nn
xx x x
xx dx C n dx x C
n x
ae dx e C a dx C
a
+
= + ≠− = ++
= + = +
∫ ∫
∫ ∫
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2 2
sin cos cos sin
sec tan csc cot
sec tan sec csc cot csc
x dx x C x dx x C
dx x C dx x C
x x dx x C x x dx x C
=− + = +
= + =− +
= + =− +
∫ ∫
∫ ∫
∫ ∫
FORMULAS OF INTEGRALS
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1 12 2 2 2
sinh cosh cosh sinh
tan ln | sec | cot ln | sin |
1 1 1tan sin
x dx x C x dx x C
x dx x C x dx x C
x xdx C dx C
x a a a aa x
− −
= + = +
= + = +
⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟ ⎜ ⎟+ ⎝ ⎠ ⎝ ⎠−
∫ ∫
∫ ∫
∫ ∫
FORMULAS OF INTEGRALS
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TECHNIQUES OF INTEGRATION
In this chapter, we develop techniques
for using the basic integration formulas.
This helps obtain indefinite integrals of more complicated functions.
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TECHNIQUES OF INTEGRATION
We learned the most important method
of integration, the Substitution Rule, in
Section 5.5
The other general technique, integration by
parts, is presented in Section 7.1
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TECHNIQUES OF INTEGRATION
Then, we learn methods that are special
to particular classes of functions—such as
trigonometric functions and rational
functions.
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TECHNIQUES OF INTEGRATION
Integration is not as straightforward
as differentiation.
There are no rules that absolutely guarantee obtaining an indefinite integral of a function.
Therefore, we discuss a strategy for integration in Section 7.5
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7.1Integration by Parts
In this section, we will learn:
How to integrate complex functions by parts.
TECHNIQUES OF INTEGRATION
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Every differentiation rule has
a corresponding integration rule.
For instance, the Substitution Rule for integration corresponds to the Chain Rule for differentiation.
INTEGRATION BY PARTS
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The rule that corresponds to
the Product Rule for differentiation
is called the rule for integration by
parts.
INTEGRATION BY PARTS
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The Product Rule states that, if f and g
are differentiable functions, then
INTEGRATION BY PARTS
[ ]( ) ( ) ( ) '( ) ( ) '( )df x g x f x g x g x f x
dx= +
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In the notation for indefinite integrals,
this equation becomes
or
INTEGRATION BY PARTS
[ ]( ) '( ) ( ) '( ) ( ) ( )f x g x g x f x dx f x g x+ =∫
( ) '( ) ( ) '( ) ( ) ( )f x g x dx g x f x dx f x g x+ =∫ ∫
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We can rearrange this equation as:
INTEGRATION BY PARTS
( ) '( ) ( ) ( ) ( ) '( )f x g x dx f x g x g x f x dx= −∫ ∫
Formula 1
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Formula 1 is called the formula for
integration by parts.
It is perhaps easier to remember in the following notation.
INTEGRATION BY PARTS
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Let u = f(x) and v = g(x).
Then, the differentials are:
du = f’(x) dx and dv = g’(x) dx
INTEGRATION BY PARTS
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INTEGRATION BY PARTS
Thus, by the Substitution Rule,
the formula for integration by parts
becomes:
u dv uv v du= −∫ ∫
Formula 2
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Find ∫ x sin x dx
Suppose we choose f(x) = x and g’(x) = sin x.
Then, f’(x) = 1 and g(x) = –cos x.
For g, we can choose any antiderivative of g’.
INTEGRATION BY PARTS E. g. 1—Solution 1
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Using Formula 1, we have:
It’s wise to check the answer by differentiating it. If we do so, we get x sin x, as expected.
INTEGRATION BY PARTS
sin ( ) ( ) ( ) '( )
( cos ) ( cos )
cos cos
cos sin
x x dx f x g x g x f x dx
x x x dx
x x x dx
x x x C
= −
= − − −
=− +
=− + +
∫ ∫∫
∫
E. g. 1—Solution 1
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Let
Then,
Using Formula 2, we have:
INTEGRATION BY PARTS
sinu x dv x dx= =
} } }sin sin ( cos ) ( cos )
cos cos
cos sin
dv v vu u du
x x dx x x dx x x x dx
x x x dx
x x x C
= = − − −
=− +
=− + +
∫ ∫ ∫∫
64 7 48 64 7 48 64 7 48
E. g. 1—Solution 2
cosdu dx v x= =−
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Our aim in using integration by parts is
to obtain a simpler integral than the one
we started with.
Thus, in Example 1, we started with ∫ x sin x dx and expressed it in terms of the simpler integral
∫ cos x dx.
NOTE
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If we had instead chosen u = sin x and
dv = x dx , then du = cos x dx and v = x2/2.
So, integration by parts gives:
Although this is true, ∫ x2cos x dx is a more difficult integral than the one we started with.
221
sin (sin ) cos2 2
xx x dx x x dx= −∫ ∫
NOTE
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Hence, when choosing u and dv, we
usually try to keep u = f(x) to be a function
that becomes simpler when differentiated.
At least, it should not be more complicated.
However, make sure that dv = g’(x) dx can be readily integrated to give v.
NOTE
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Evaluate ∫ ln x dx
Here, we don’t have much choice for u and dv.
Let
Then,
INTEGRATION BY PARTS Example 2
lnu x dv dx= =
1du dx v x
x= =
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Integrating by parts, we get:
INTEGRATION BY PARTS
ln ln
ln
ln
dxx dx x x x
x
x x dx
x x x C
= −
= −
= − +
∫ ∫∫
Example 2
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INTEGRATION BY PARTS
Integration by parts is effective in
this example because the derivative of
the function f(x) = ln x is simpler than f.
Example 2
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Find ∫ t2etdt
Notice that t2 becomes simpler when differentiated.
However, et is unchanged when differentiated or integrated.
INTEGRATION BY PARTS Example 3
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INTEGRATION BY PARTS
So, we choose
Then,
Integration by parts gives:
2 tu t dv e dt= =
2 tdu t dt v e= =
E. g. 3—Equation 3
2 2 2t t tt e dt t e te dt= −∫ ∫
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INTEGRATION BY PARTS Example 3
The integral that we obtained, ∫ tetdt,
is simpler than the original integral.
However, it is still not obvious.
So, we use integration by parts a second time.
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INTEGRATION BY PARTS
This time, we choose
u = t and dv = etdt
Then, du = dt, v = et.
So,t t t t tte dt te e dt te e C= − − − +∫ ∫
Example 3
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Putting this in Equation 3, we get
where C1 = – 2C
INTEGRATION BY PARTS
2 2
2
21
2
2( )
2 2
t t t
t t t
t t t
t e dt t e te dt
t e te e C
t e te e C
= −
= − − +
= − − +
∫ ∫
Example 3
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Evaluate ∫ ex sinx dx
ex does not become simpler when differentiated.
Neither does sin x become simpler.
INTEGRATION BY PARTS Example 4
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INTEGRATION BY PARTS
Nevertheless, we try choosing
u = ex and dv = sin x
Then, du = ex dx and v = – cos x.
E. g. 4—Equation 4
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INTEGRATION BY PARTS
So, integration by parts gives:
sin cos cosx x xe x dx e x e x dx=− +∫ ∫
Example 4
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The integral we have obtained, ∫ excos x dx,
is no simpler than the original one.
At least, it’s no more difficult.
Having had success in the preceding example integrating by parts twice, we do it again.
INTEGRATION BY PARTS Example 4
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This time, we use
u = ex and dv = cos x dx
Then, du = ex dx, v = sin x, and
INTEGRATION BY PARTS
cos sin sinx x xe x dx e x e x dx= −∫ ∫
E. g. 4—Equation 5
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At first glance, it appears as if we have
accomplished nothing.
We have arrived at ∫ ex sin x dx, which is where we started.
INTEGRATION BY PARTS Example 4
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However, if we put the expression for
∫ ex cos x dx from Equation 5 into Equation 4,
we get:
This can be regarded as an equation to be solved for the unknown integral.
INTEGRATION BY PARTS Example 4
sin cos sin
sin
x x x
x
e x dx e x e x
e x dx
=− +
−
∫∫
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Adding to both sides ∫ ex sin x dx,
we obtain:
INTEGRATION BY PARTS Example 4
2 sin cos sinx x xe x dx e x e x=− +∫
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Dividing by 2 and adding the constant
of integration, we get:
INTEGRATION BY PARTS Example 4
12sin (sin cos )x xe x dx e x x C= − +∫
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The figure illustrates the example by
showing the graphs of f(x) = ex sin x and
F(x) = ½ ex(sin x – cos x).
As a visual check on our work, notice that f(x) = 0 when F has a maximum or minimum.
INTEGRATION BY PARTS
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If we combine the formula for integration
by parts with Part 2 of the FTC (FTC2),
we can evaluate definite integrals by parts.
INTEGRATION BY PARTS
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Evaluating both sides of Formula 1 between
a and b, assuming f’ and g’ are continuous,
and using the FTC, we obtain:
INTEGRATION BY PARTS
]( ) '( ) ( ) ( )
( ) '( )
b b
aa
b
a
f x g x dx f x g x
g x f x dx
=
−
∫
∫
Formula 6
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Calculate
Let
Then,
INTEGRATION BY PARTS Example 51 1
0tan x dx−∫
1tanu x dv dx−= =
21
dxdu v x
x= =
+
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So, Formula 6 gives:
INTEGRATION BY PARTS Example 5
1 111 1200 0
11 120
1
20
tan tan1
1 tan 1 0 tan 01
4 1
xx dx x x dx
xxdx
xxdx
x
π
− −
− −
⎤= −⎦ +
= ⋅ − ⋅ −+
= −+
∫ ∫
∫
∫
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To evaluate this integral, we use
the substitution t = 1 + x2 (since u has
another meaning in this example).
Then, dt = 2x dx.
So, x dx = ½ dt.
INTEGRATION BY PARTS Example 5
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When x = 0, t = 1, and when x = 1, t = 2.
Hence,
INTEGRATION BY PARTS
]
1 21220 1
212 1
12
12
1
ln | |
(ln 2 ln1)
ln 2
x dtdx
x t
t
=+
=
= −
=
∫ ∫
Example 5
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INTEGRATION BY PARTS
Therefore,
1 1120 0
tan4 1
ln 2
4 2
xx dx dx
x
π
π
− = −+
= −
∫ ∫
Example 5
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As tan-1x ≥ for x ≥ 0 , the integral in
the example can be interpreted as the area
of the region shown here.
INTEGRATION BY PARTS
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Prove the reduction formula
where n ≥ 2 is an integer.
This is called a reduction formula because the exponent n has been reduced to n – 1 and n – 2.
INTEGRATION BY PARTS E. g. 6—Formula 7
1
2
1sin cos sin
1sin
n n
n
x dx x xn
nx dx
n
−
−
=−
−+
∫
∫
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Let
Then,
So, integration by parts gives:
INTEGRATION BY PARTS
1sin sinnu x dv x dx−= =
1
2 2
sin cos sin
( 1) sin cos
n n
n
x dx x x
n x x dx
−
−
=−
+ −
∫∫
Example 6
2( 1)sin cos cosndu n x x dx v x−= − =−
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Since cos2x = 1 – sin2x, we have:
As in Example 4, we solve this equation for the desired integral by taking the last term on the right side to the left side.
INTEGRATION BY PARTS Example 6
1 2sin cos sin ( 1) sin
( 1) sin
n n n
n
x dx x x n x dx
n x dx
− −=− + −
− −
∫ ∫∫
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Thus, we have:
or
INTEGRATION BY PARTS Example 6
1 2sin cos sin ( 1) sinn n nn x dx x x n x dx− −=− + −∫ ∫
1 21 ( 1)sin cos sin sinn n nn
x dx x x x dxn n
− −−=− +∫ ∫
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The reduction formula (7) is useful.
By using it repeatedly, we could express
∫ sinnx dx in terms of:
∫ sin x dx (if n is odd)
∫ (sin x)0dx = ∫ dx (if n is even)
INTEGRATION BY PARTS