8 Techniques of Integration

8.3 Trigonometric Substitution

In Section 5.2 we defined the definite integral as the “signed area under the curve.” In that section we had not yet learned the Fundamental Theorem of Calculus, so we evaluated special definite integrals which described nice, geometric shapes. For instance, we were able to evaluate

-339-x2dx=9π2 (8.3.1)

as we recognized that f(x)=9-x2 described the upper half of a circle with radius 3.

We have since learned a number of integration techniques, including Substitution and Integration by Parts, yet we are still unable to evaluate the above integral without resorting to a geometric interpretation. This section introduces Trigonometric Substitution, a method of integration that fills this gap in our integration skill. This technique works on the same principle as Substitution as found in Section 5.5, though it can feel “backward.” In Section 5.5, we set u=f(x), for some function f, and replaced f(x) with u. In this section, we will set x=f(θ), where f is a trigonometric function, then replace x with f(θ).

We start by demonstrating this method in evaluating the integral in Equation (8.3.1). After the example, we will generalize the method and give more examples.

Example 8.3.1 Using Trigonometric Substitution

Evaluate -339-x2dx.

SolutionWe begin by noting that 9sin2θ+9cos2θ=9, and hence 9cos2θ=9-9sin2θ. If we let x=3sinθ, then 9-x2=9-9sin2θ=9cos2θ.

Setting x=3sinθ gives dx=3cosθdθ. We are almost ready to substitute. We also change our bounds of integration. The bound x=-3 corresponds to θ=-π/2 (for when θ=-π/2, x=3sinθ=-3). Likewise, the bound of x=3 is replaced by the bound θ=π/2. Thus

-339-x2dx =-π/2π/29-9sin2θ(3cosθ)dθ
On [-π/2,π/2], cosθ is always positive, so we can drop the absolute value bars, then employ a half-angle formula:

This matches our answer from before.

We now describe in detail Trigonometric Substitution. This method excels when dealing with integrands that contain a2-x2, x2-a2 and x2+a2. The following Key Idea outlines the procedure for each case, followed by more examples.

Key Idea 8.3.1 Trigonometric Substitution

  1. (a)

    For integrands containing a2-x2:
    Let x=asinθ,  for -π/2θπ/2 and a>0.
    On this interval, cosθ0, so a2-x2=acosθ

  2. (b)

    For integrands containing x2+a2:
    Let x=atanθ,  for -π/2<θ<π/2 and a>0.
    On this interval, secθ>0, so x2+a2=asecθ

  3. (c)

    For integrands containing x2-a2:
    Let x=asecθ,  restricting our work to where xa>0,
    so x/a1, and 0θ<π/2.
    On this interval, tanθ0, so x2-a2=atanθ

Example 8.3.2 Using Trigonometric Substitution

Evaluate 15+x2dx.

SolutionUsing Key Idea 8.3.1(b), we recognize a=5 and set x=5tanθ. This makes dx=5sec2θdθ. We will use the fact that 5+x2=5+5tan2θ=5sec2θ=5secθ. Substituting, we have:

15+x2dx =15+5tan2θ5sec2θdθ

While the integration steps are over, we are not yet done. The original problem was stated in terms of x, whereas our answer is given in terms of θ. We must convert back to x.

margin: 5xx2+5θ Figure 8.3.1: A reference triangle for Example 8.3.2 Λ

The lengths of the sides of the reference triangle in Figure 8.3.1 are determined by the Pythagorean Theorem. With x=5tanθ, we have


This gives

15+x2dx =ln|secθ+tanθ|+C

We can leave this answer as is, or we can use a logarithmic identity to simplify it. Note:

ln|x2+55+x5|+C =ln|15(x2+5+x)|+C

where the ln(1/5) term is absorbed into the constant C. (In Section 7.4 we learned another way of approaching this problem.)

Example 8.3.3 Using Trigonometric Substitution

Evaluate 4x2-1dx.

SolutionWe start by rewriting the integrand so that it has the form x2-a2 for some value of a:

4x2-1 =4(x2-14)

So we have a=1/2, and following Key Idea 8.3.1(c), we set x=12secθ, and hence dx=12secθtanθdθ. We now rewrite the integral with these substitutions:

4x2-1dx =2x2-(12)2dx

We integrated sec3θ in Example 8.2.6, finding its antiderivatives to be




We are not yet done. Our original integral is given in terms of x, whereas our final answer, as given, is in terms of θ. We need to rewrite our answer in terms of x. With a=1/2, and x=12secθ, we use the Pythagorean Theorem to determine the lengths of the sides of the reference triangle in Figure 8.3.2. margin: 1/2x2-1/4xθ Figure 8.3.2: A reference triangle for Example 8.3.3 Λ

tanθ=x2-1412=2x2-14  and  secθ=2x.


4x2-1dx =14(secθtanθ-ln|secθ+tanθ|)+C
Example 8.3.4 Using Trigonometric Substitution

Evaluate 4-x2x2dx.

SolutionWe use Key Idea 8.3.1(a) with a=2, x=2sinθ, dx=2cosθdθ and hence 4-x2=2cosθ. This gives

4-x2x2dx =2cosθ4sin2θ(2cosθ)dθ
margin: 4-x2x2θ Figure 8.3.3: A reference triangle for Example 8.3.4 Λ

We need to rewrite our answer in terms of x. Using the Pythagorean Theorem we determine the lengths of the sides of the reference triangle in Figure 8.3.3. We have cotθ=4-x2/x and θ=sin-1(x/2). Thus


Trigonometric Substitution can be applied in many situations, even those not of the form a2-x2, x2-a2 or x2+a2. In the following example, we apply it to an integral we already know how to handle.

Example 8.3.5 Using Trigonometric Substitution

Evaluate 1x2+1dx.

SolutionWe know the answer already as tan-1x+C. We apply Trigonometric Substitution here to show that we get the same answer without inherently relying on knowledge of the derivative of the arctangent function.

Using Key Idea 8.3.1(b), let x=tanθ, dx=sec2θdθ and note that x2+1=tan2θ+1=sec2θ. Thus

1x2+1dx =1sec2θsec2θdθ

Since x=tanθ, θ=tan-1x, and we conclude that 1x2+1dx=tan-1x+C.

The next example is similar to the previous one in that it does not involve a square-root. It shows how several techniques and identities can be combined to obtain a solution.

Example 8.3.6 Using Trigonometric Substitution

Evaluate 1(x2+6x+10)2dx.

SolutionWe start by completing the square, then make the substitution u=x+3, followed by the trigonometric substitution of u=tanθ: margin: Note: Remember the sine and cosine double angle identities: sin2θ =2sinθcosθ cos2θ =cos2θ-sin2θ =2cos2θ-1 =1-2sin2θ They are often needed for writing your final answer in terms of x. Λ

1(x2+6x+10)2dx =1((x+3)2+1)2dx=1(u2+1)2du.
Now make the substitution u=tanθ, du=sec2θdθ:
Applying a half-angle formula, we have
=12θ+14sin(2θ)+C. (8.3.2)

We need to return to the variable x. As u=tanθ, θ=tan-1u. Using the identity sin(2θ)=2sinθcosθ and using the reference triangle found in Key Idea 8.3.1(b), we have


Finally, we return to x with the substitution u=x+3. We start with the expression in Equation (8.3.2):

12θ+14sin(2θ)+C =12tan-1u+12uu2+1+C

Stating our final result in one line,


Our last example returns us to definite integrals, as seen in our first example. Given a definite integral that can be evaluated using Trigonometric Substitution, we could first evaluate the corresponding indefinite integral (by changing from an integral in terms of x to one in terms of θ, then converting back to x) and then evaluate using the original bounds. It is much more straightforward, though, to change the bounds as we substitute.

Example 8.3.7 Definite integration and Trigonometric Substitution

Evaluate 05x2x2+25dx.

SolutionUsing Key Idea 8.3.1(b), we set x=5tanθ, dx=5sec2θdθ, and note that x2+25=5secθ. As we substitute, we change the bounds of integration.

The lower bound of the original integral is x=0. As x=5tanθ, we solve for θ and find θ=tan-1(x/5). Thus the new lower bound is θ=tan-1(0)=0. The original upper bound is x=5, thus the new upper bound is θ=tan-1(5/5)=π/4.

Thus we have

05x2x2+25dx =0π/425tan2θ5secθ5sec2θdθ

We encountered this indefinite integral in Example 8.3.3 where we found



250π/4tan2θsecθdθ =252(secθtanθ-ln|secθ+tanθ|)|0π/4

The next section introduces Partial Fraction Decomposition, which is an algebraic technique that turns “complicated” fractions into sums of “simpler” fractions, making integration easier.

Exercises 8.3


Terms and Concepts

  1. 1.

    Trigonometric Substitution works on the same principles as Integration by Substitution, though it can feel “            ”.

  2. 2.

    If one uses Trigonometric Substitution on an integrand containing 25-x2, then one should set x=            .

  3. 3.
    Consider the Pythagorean Identity sin2θ+cos2θ=1. (a) What identity is obtained when both sides are divided by cos2θ? (b) Use the new identity to simplify 9tan2θ+9.
  4. 4.

    Why does Key Idea 8.3.1(a) state that a2-x2=acosθ, and not |acosθ|?


In Exercises 5–26., apply Trigonometric Substitution to evaluate the indefinite integrals.

  1. 5.


  2. 6.


  3. 7.


  4. 8.


  5. 9.


  6. 10.


  7. 11.


  8. 12.


  9. 13.


  10. 14.


  11. 15.


  12. 16.


  13. 17.


  14. 18.


  15. 19.


  16. 20.


  17. 21.


  18. 22.


  19. 23.


  20. 24.


  21. 25.


  22. 26.


In Exercises 27–34., evaluate the indefinite integrals. Some may be evaluated without Trigonometric Substitution.

  1. 27.


  2. 28.


  3. 29.


  4. 30.


  5. 31.


  6. 32.


  7. 33.


  8. 34.


In Exercises 35–40., evaluate the definite integrals by making the proper trigonometric substitution and changing the bounds of integration.

  1. 35.


  2. 36.


  3. 37.


  4. 38.


  5. 39.


  6. 40.


  1. 41.
    Find the volume of the solid of revolution obtained by rotating the region bounded by y=0, y=x1+x2, x=0, and x=1: (a) About the x-axis, using the disk method. (b) About the y-axis, using the shell method.
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