TECHNIQUES OF INTEGRATION

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TECHNIQUES OF INTEGRATION
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TECHNIQUES OF INTEGRATION

• 7.3
• Trigonometric Substitution

In this section, we will learn about The various
types of trigonometric substitutions.
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TRIGONOMETRIC SUBSTITUTION

• In finding the area of a circle or an ellipse,
  an integral of the form
  arises, where a gt 0.
• If it were , the
  substitution would be effective.
• However, as it stands, is
  more difficult.

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TRIGONOMETRIC SUBSTITUTION

• If we change the variable from x to ? by the
  substitution x a sin ?, the identity 1 sin2?
  cos2? lets us lose the root sign.
• This is because

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TRIGONOMETRIC SUBSTITUTION

• Notice the difference between the substitution u
  a2 x2 and the substitution x a sin ?.
• In the first, the new variable is a function of
  the old one.
• In the second, the old variable is a function of
  the new one.

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TRIGONOMETRIC SUBSTITUTION

• In general, we can make a substitution of the
  form x g(t) by using the Substitution Rule
  in reverse.
• To make our calculations simpler, we assume g
  has an inverse function, that is, g is
  one-to-one.

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INVERSE SUBSTITUTION

• Here, if we replace u by x and x by t in the
  Substitution Rule (Equation 4 in Section 5.5), we
  obtain
• This kind of substitution is called inverse
  substitution.

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INVERSE SUBSTITUTION

• We can make the inverse substitution x a sin
  ?, provided that it defines a one-to-one
  function.
• This can be accomplished by restricting ? to lie
  in the interval -p/2, p/2.

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TABLE OF TRIGONOMETRIC SUBSTITUTIONS

• Here, we list trigonometric substitutions that
  are effective for the given radical expressions
  because of the specified trigonometric identities.

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TABLE OF TRIGONOMETRIC SUBSTITUTIONS

• In each case, the restriction on ? is imposed to
  ensure that the function that defines the
  substitution is one-to-one.
• These are the same intervals used in Section 1.6
  in defining the inverse functions.

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TRIGONOMETRIC SUBSTITUTION
Example 1

• Evaluate
• Let x 3 sin ?, where p/2 ? p/2.
• Then, dx 3 cos ? d? and
• Note that cos ? 0 because p/2 ? p/2.)

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TRIGONOMETRIC SUBSTITUTION
Example 1

• Thus, the Inverse Substitution Rule gives

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TRIGONOMETRIC SUBSTITUTION
Example 1

• As this is an indefinite integral, we must
  return to the original variable x.
• This can be done in either of two ways.

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TRIGONOMETRIC SUBSTITUTION
Example 1

• One, we can use trigonometric identities to
  express cot ? in terms of sin ? x/3.

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TRIGONOMETRIC SUBSTITUTION
Example 1

• Two, we can draw a diagram, where ? is
  interpreted as an angle of a right triangle.

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TRIGONOMETRIC SUBSTITUTION
Example 1

• Since sin ? x/3, we label the opposite side and
  the hypotenuse as having lengths x and 3.

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TRIGONOMETRIC SUBSTITUTION
Example 1

• Then, the Pythagorean Theorem gives the length
  of the adjacent side as

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TRIGONOMETRIC SUBSTITUTION
Example 1

• So, we can simply read the value of cot ? from
  the figure
• Although ? gt 0 here, this expression for cot ?
  is valid even when ? lt 0.

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TRIGONOMETRIC SUBSTITUTION
Example 1

• As sin ? x/3, we have ? sin-1(x/3).
• Hence,

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TRIGONOMETRIC SUBSTITUTION
Example 2

• Find the area enclosed by the ellipse

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TRIGONOMETRIC SUBSTITUTION
Example 2

• Solving the equation of the ellipse for y, we
  get
• or

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TRIGONOMETRIC SUBSTITUTION
Example 2

• As the ellipse is symmetric with respect to both
  axes, the total area A is four times the area in
  the first quadrant.

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TRIGONOMETRIC SUBSTITUTION
Example 2

• The part of the ellipse in the first quadrant is
  given by the function
• Hence,

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TRIGONOMETRIC SUBSTITUTION
Example 2

• To evaluate this integral, we substitute x a
  sin ?.
• Then, dx a cos ? d?.

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TRIGONOMETRIC SUBSTITUTION
Example 2

• To change the limits of integration, we note
  that
• When x 0, sin ? 0 so ? 0
• When x a, sin ? 1 so ? p/2

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TRIGONOMETRIC SUBSTITUTION
Example 2

• Also, since 0 ? p/2,

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TRIGONOMETRIC SUBSTITUTION
Example 2

• Therefore,

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TRIGONOMETRIC SUBSTITUTION
Example 2

• We have shown that the area of an ellipse with
  semiaxes a and b is pab.
• In particular, taking a b r, we have proved
  the famous formula that the area of a circle
  with radius r is pr2.

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TRIGONOMETRIC SUBSTITUTION
Note

• The integral in Example 2 was a definite
  integral.
• So, we changed the limits of integration, and
  did not have to convert back to the original
  variable x.

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TRIGONOMETRIC SUBSTITUTION
Example 3

• Find
• Let x 2 tan ?, p/2 lt ? lt p/2.
• Then, dx 2 sec2 ? d? and

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TRIGONOMETRIC SUBSTITUTION
Example 3

• Thus, we have

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TRIGONOMETRIC SUBSTITUTION
Example 3

• To evaluate this trigonometric integral, we put
  everything in terms of sin ? and cos ?

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TRIGONOMETRIC SUBSTITUTION
Example 3zz

• Therefore, making the substitution u sin ?, we
  have

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TRIGONOMETRIC SUBSTITUTION
Example 3

• We use the figure to determine that
• Hence,

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TRIGONOMETRIC SUBSTITUTION
Example 4

• Find
• It would be possible to use the trigonometric
  substitution x 2 tan ? (as in Example 3).

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TRIGONOMETRIC SUBSTITUTION
Example 4

• However, the direct substitution u x2 4 is
  simpler.
• This is because, then, du 2x dx and

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TRIGONOMETRIC SUBSTITUTION
Note

• Example 4 illustrates the fact that, even when
  trigonometric substitutions are possible, they
  may not give the easiest solution.
• You should look for a simpler method first.

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TRIGONOMETRIC SUBSTITUTION
Example 5

• Evaluate where a gt 0.

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TRIGONOMETRIC SUBSTITUTION
E. g. 5Solution 1

• We let x a sec ?, where 0 lt ? lt p/2 or p lt ? lt
  p/2.
• Then, dx a sec ? tan ? d? and

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TRIGONOMETRIC SUBSTITUTION
E. g. 5Solution 1

• Therefore,

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TRIGONOMETRIC SUBSTITUTION
E. g. 5Solution 1

• The triangle in the figure gives

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TRIGONOMETRIC SUBSTITUTION
E. g. 5Solution 1

• So, we have

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TRIGONOMETRIC SUBSTITUTION
E. g. 5Sol. 1 (For. 1)

• Writing C1 C ln a, we have

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TRIGONOMETRIC SUBSTITUTION
E. g. 5Solution 2

• For x gt 0, the hyperbolic substitution x a
  cosh t can also be used.
• Using the identity cosh2y sinh2y 1, we have

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TRIGONOMETRIC SUBSTITUTION
E. g. 5Solution 2

• Since dx a sinh t dt, we obtain

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TRIGONOMETRIC SUBSTITUTION
E. g. 5Sol. 2 (For. 2)

• Since cosh t x/a, we have t cosh-1(x/a) and

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TRIGONOMETRIC SUBSTITUTION
E. g. 5Sol. 2 (For. 2)

• Although Formulas 1 and 2 look quite different,
  they are actually equivalent by Formula 4 in
  Section 3.11

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TRIGONOMETRIC SUBSTITUTION
Note

• As Example 5 illustrates, hyperbolic
  substitutions can be used instead of
  trigonometric substitutions, and sometimes they
  lead to simpler answers.
• However, we usually use trigonometric
  substitutions, because trigonometric identities
  are more familiar than hyperbolic identities.

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TRIGONOMETRIC SUBSTITUTION
Example 6

• Find
• First, we note that
• So, trigonometric substitution is appropriate.

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TRIGONOMETRIC SUBSTITUTION
Example 6

• is not quite one of the
  expressions in the table of trigonometric
  substitutions.
• However, it becomes one if we make the
  preliminary substitution u 2x.

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TRIGONOMETRIC SUBSTITUTION
Example 6

• When we combine this with the tangent
  substitution, we have .
• This gives and

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TRIGONOMETRIC SUBSTITUTION
Example 6

• When x 0, tan ? 0 so ? 0.
• When x , tan ? so ? p/3.

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TRIGONOMETRIC SUBSTITUTION
Example 6
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TRIGONOMETRIC SUBSTITUTION
Example 6

• Now, we substitute u cos ? so that du - sin
  ? d?.
• When ? 0, u 1.
• When ? p/3, u ½.

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TRIGONOMETRIC SUBSTITUTION
Example 6

• Therefore,

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TRIGONOMETRIC SUBSTITUTION
Example 7

• Evaluate
• We can transform the integrand into a function
  for which trigonometric substitution is
  appropriate, by first completing the square
  under the root sign

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TRIGONOMETRIC SUBSTITUTION
Example 7

• This suggests we make the substitution u x
  1.
• Then, du dx and x u 1.
• So,

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TRIGONOMETRIC SUBSTITUTION
Example 7

• We now substitute .
• This gives and

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TRIGONOMETRIC SUBSTITUTION
Example 7

• So,

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TRIGONOMETRIC SUBSTITUTION

• The figure shows the graphs of the integrand in
  Example 7 and its indefinite integral (with C
  0).
• Which is which?