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  • Solve integration problems involving the square root of a sum or difference of two squares.

In this section, we explore integrals containing expressions of the form a 2 x 2 , a 2 + x 2 , and x 2 a 2 , where the values of a are positive. We have already encountered and evaluated integrals containing some expressions of this type, but many still remain inaccessible. The technique of trigonometric substitution    comes in very handy when evaluating these integrals. This technique uses substitution to rewrite these integrals as trigonometric integrals.

Integrals involving a 2 x 2

Before developing a general strategy for integrals containing a 2 x 2 , consider the integral 9 x 2 d x . This integral cannot be evaluated using any of the techniques we have discussed so far. However, if we make the substitution x = 3 sin θ , we have d x = 3 cos θ d θ . After substituting into the integral, we have

9 x 2 d x = 9 ( 3 sin θ ) 2 3 cos θ d θ .

After simplifying, we have

9 x 2 d x = 9 1 sin 2 θ cos θ d θ .

Letting 1 sin 2 θ = cos 2 θ , we now have

9 x 2 d x = 9 cos 2 θ cos θ d θ .

Assuming that cos θ 0 , we have

9 x 2 d x = 9 cos 2 θ d θ .

At this point, we can evaluate the integral using the techniques developed for integrating powers and products of trigonometric functions. Before completing this example, let’s take a look at the general theory behind this idea.

To evaluate integrals involving a 2 x 2 , we make the substitution x = a sin θ and d x = a cos θ . To see that this actually makes sense, consider the following argument: The domain of a 2 x 2 is [ a , a ] . Thus, a x a . Consequently, −1 x a 1 . Since the range of sin x over [ ( π / 2 ) , π / 2 ] is [ −1 , 1 ] , there is a unique angle θ satisfying ( π / 2 ) θ π / 2 so that sin θ = x / a , or equivalently, so that x = a sin θ . If we substitute x = a sin θ into a 2 x 2 , we get

a 2 x 2 = a 2 ( a sin θ ) 2 Let x = a sin θ where π 2 θ π 2 . Simplify. = a 2 a 2 sin 2 θ Factor out a 2 . = a 2 ( 1 sin 2 θ ) Substitute 1 sin 2 x = cos 2 x . = a 2 cos 2 θ Take the square root. = | a cos θ | = a cos θ .

Since cos x 0 on π 2 θ π 2 and a > 0 , | a cos θ | = a cos θ . We can see, from this discussion, that by making the substitution x = a sin θ , we are able to convert an integral involving a radical into an integral involving trigonometric functions. After we evaluate the integral, we can convert the solution back to an expression involving x . To see how to do this, let’s begin by assuming that 0 < x < a . In this case, 0 < θ < π 2 . Since sin θ = x a , we can draw the reference triangle in [link] to assist in expressing the values of cos θ , tan θ , and the remaining trigonometric functions in terms of x . It can be shown that this triangle actually produces the correct values of the trigonometric functions evaluated at θ for all θ satisfying π 2 θ π 2 . It is useful to observe that the expression a 2 x 2 actually appears as the length of one side of the triangle. Last, should θ appear by itself, we use θ = sin −1 ( x a ) .

This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled a, the vertical leg is labeled x, and the horizontal leg is labeled as the square root of (a^2 – x^2). To the left of the triangle is the equation sin(theta) = x/a.
A reference triangle can help express the trigonometric functions evaluated at θ in terms of x .

The essential part of this discussion is summarized in the following problem-solving strategy.

Problem-solving strategy: integrating expressions involving a 2 x 2

  1. It is a good idea to make sure the integral cannot be evaluated easily in another way. For example, although this method can be applied to integrals of the form 1 a 2 x 2 d x , x a 2 x 2 d x , and x a 2 x 2 d x , they can each be integrated directly either by formula or by a simple u -substitution.
  2. Make the substitution x = a sin θ and d x = a cos θ d θ . Note : This substitution yields a 2 x 2 = a cos θ .
  3. Simplify the expression.
  4. Evaluate the integral using techniques from the section on trigonometric integrals.
  5. Use the reference triangle from [link] to rewrite the result in terms of x . You may also need to use some trigonometric identities and the relationship θ = sin −1 ( x a ) .
Practice Key Terms 1

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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