LESSON 9 INTEGRATION BY TRIGONOMETRIC SUBSTITUTION
If the integrand of an integral contains an expression of the form , where a is a positive real number and n is an odd integer, then the domain of the integrand is either the closed interval or is a subset of this interval. Then try the trigonometric substitution: Let , where . Then . Note that the range of the function is the closed interval . Also, we have that = = = = . Thus, = = =
= . Since , then . Since whenever , then . Thus, = .
If , then . Using right triangle trigonometry, we have that
a
x
, , ,
If the integrand of an integral contains an expression of the form , where a is a positive real number and n is an integer, then the domain of the integrand is the set of all real numbers. Then try the trigonometric substitution: Let , where . Then . Note that the range of the function is the set of all real numbers. Also, we have that = = = = . Thus, = = = = . Since , then . Since whenever , then . Thus, = .
If , then . Using right triangle trigonometry, we have that
x
a
, , ,
If the integrand of an integral contains an expression of the form , where a is a positive real number and n is an odd integer, then the domain of the integrand is the set or is a subset of this set. Then try the trigonometric substitution: Let , where or . Then . Note that the range of the function is the set . Also, we have that = = = = . Thus, = = = = . Since , then . Since whenever or , then . Thus, = .
If , then . Using right triangle trigonometry, we have that
x
a
, , ,
Examples Evaluate the following integrals.
1.
Note that we do not have the x in the integral to do the simple substitution to let . We will use the technique of Trigonometric Substitution.
Let , where . Then .
Also, = = .
Thus, since when . Thus, we have that
= = =
Since , then . Using right triangle trigonometry, we have that .
x
3
Thus, = = =
= =
= , where
Answer:
2.
Since the function is continuous on its domain of definition, which is the open interval , then it is continuous on the closed interval . Thus, the Fundamental Theorem of Calculus can be applied.
We will first evaluate the indefinite integral using the technique of Trigonometric Substitution. Then we will apply the Fundamental Theorem of Calculus with one of these antiderivatives.
Let , where . Then .
Also, = = .
Thus, since when . Thus, we have that
= = =
=
Let . Then
= =
= = =
=
Since , then . Using right triangle trigonometry, we have that .
5
x
Thus, = = =
= =
=
Now, that we have all the antiderivatives of the integrand , we can apply the Fundamental Theorem of Calculus to evaluate the definite integral .
= =
= = =
Answer:
3.
Note that we do not have the t in the integral to do the simple substitution to let .
Since the function is continuous on its domain of definition, which is the closed interval , then it is continuous on the closed interval . Thus, the Fundamental Theorem of Calculus can be applied.
However, the value of this definite integral can be obtained using the area of the region bounded by the graph of the integrand and the horizontal t-axis on the closed interval . Since the graph of the function is the top-half of the circle given by the equation . The center of the circle is the origin and the radius is 7. Thus, the region bounded by the graph of and the t-axis on the interval is the upper left quarter of the circle. Since the area of the circle is given by and , then the area of the circle is . Thus, the area of the quarter circle is . Thus, = .
To apply the Fundamental Theorem of Calculus, we will first evaluate the indefinite integral using the technique of Trigonometric Substitution in order to obtain all the antiderivatives of the integrand .
Let , where . Then .
Also, = = .
Thus, since when . Thus, we have that
= = =
Since , then and . Using right triangle trigonometry, we have that .
7
t
By the double angle formula for sine, we have that .
Thus, = .
Thus, = = =
=
=
Now, that we have all the antiderivatives of the integrand , we can apply the Fundamental Theorem of Calculus to evaluate the definite integral .
= =
= =
= =
Answer:
4.
Since the function is continuous on its domain of definition, which is the set , then it is continuous on the closed interval . Thus, the Fundamental Theorem of Calculus can be applied.
We will first evaluate the indefinite integral using the technique of Trigonometric Substitution. Then we will apply the Fundamental Theorem of Calculus with one of these antiderivatives.
Let , where or . Then .
Also, = = .
Thus, since when or . Thus, we have that
= = =
=
Let . Then
= = =
=
Since , then . Using right triangle trigonometry, we have that .
x
4
Thus, = = =
= =
=
Now, that we have all the antiderivatives of the integrand , we can apply the Fundamental Theorem of Calculus to evaluate the definite integral .
= =
= = =
Answer:
5.
We want to rewrite the integral before we use the technique of Trigonometric Substitution.
= =
=
Let , where . Then .
Also, = = .
Thus, since when . Thus, we have that
= =
=
= =
= =
=
Let . Then
= =
= =
=
Since , then . Using right triangle trigonometry, we have that .
3
Thus, = =
= =
=
=
=
Answer:
6.
This integral can be evaluated by simple substitution which is FASTER than trigonometric substitution.
Let
Then
= = =
= = =
If you missed the simple substitution and proceeded with the technique of Trigonometric Substitution, then your work would have been the following (and the following, and the following.)
= =
Let , where or . Then .
Also, = = .
Thus, since when or . Thus, we have that
= =
=
Let . Then
= = =
Since , then . Using right triangle trigonometry, we have that .
2 x
Thus, = =
= =
=
Answer:
7.
Let , where . Then .
Also, = = .
Thus, since when . Thus, we have that
= =
= =
Let . Then
=
= =
= =
=
Since , then . Using right triangle trigonometry, we have that .
t
Thus, = =
=
=
Answer:
8.
Note that we do not have the y in the integral to do the simple substitution to let . We will use the technique of Trigonometric Substitution. First, we will need to rewrite the integral.
= = =
Let , where or . Then .
Also, = = .
Thus, since when or . Thus, we have that
= =
= =
= =
Let . Then
= = = =
= =
Since , then . Using right triangle trigonometry, we have that .
3 y
5
Thus, = = =
=
Answer:
9.
= =
= =
If you evaluate the integral using the technique of Trigonometric Substitution, then your work would have been the following.
Let , where . Then .
Also, = = .
Thus, . Thus, we have that
= = =
=
Let . Then
= =
= =
=
Since , then .
Thus, = =
=
=
=
=
, where
Answer:
Example If a is a positive real number, b is any real number, m is a rational number, and n is an odd integer, then determine the trigonometric integral for the integral if you do the trigonometric substitution to let , where .
Since , then .
= = = = .
Thus, = = = =
. Since , then . Since whenever , then . Thus, = .
Thus, = =
.
Answer: =
For example, in our Example 2 above, we had the integral .
For this integral, , , , and . Thus, =
= .
Example If a is a positive real number, b is any real number, m is a rational number, and n is an integer, then determine the trigonometric integral for the integral if you do the trigonometric substitution to let , where .
Since , then .
= = = = .
Thus, = = = =
. Since , then . Since whenever , then . Thus, = .
Thus, = =
.
Answer: =
For example, in our Example 1 above, we had the integral . For this integral, , , , and . Thus, =
= .
Example If a is a positive real number, b is any real number, m is a rational number, and n is an odd integer, then determine the trigonometric integral for the integral if you do the trigonometric substitution to let , where or .
Since , then .
= = = = .
Thus, = = = =
. Since , then . Since whenever or , then . Thus, = .
Thus, = =
.
Answer: =
For example, in our Example 8 above, we had the integral =
= = . For the integral , , , , and . Thus, = = = =
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860