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Problem-solving strategy: integrals involving x 2 a 2

  1. Check to see whether the integral cannot be evaluated using another method. If so, we may wish to consider applying an alternative technique.
  2. Substitute x = a sec θ and d x = a sec θ tan θ d θ . This substitution yields
    x 2 a 2 = ( a sec θ ) 2 a 2 = a 2 ( sec 2 θ + 1 ) = a 2 tan 2 θ = | a tan θ | .

    For x a , | a tan θ | = a tan θ and for x a , | a tan θ | = a tan θ .
  3. Simplify the expression.
  4. Evaluate the integral using techniques from the section on trigonometric integrals.
  5. Use the reference triangles from [link] to rewrite the result in terms of x . You may also need to use some trigonometric identities and the relationship θ = sec −1 ( x a ) . ( Note : We need both reference triangles, since the values of some of the trigonometric ratios are different depending on whether x > a or x < a . )
This figure has two right triangles. The first triangle is in the first quadrant of the xy coordinate system and has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled x, the vertical leg is labeled the square root of (x^2-a^2), and the horizontal leg is labeled a. The horizontal leg is on the x-axis. To the left of the triangle is the equation sec(theta) = x/a, x>a. There are also the equations sin(theta)= the square root of (x^2-a^2)/x, cos(theta) = a/x, and tan(theta) = the square root of (x^2-a^2)/a. The second triangle is in the second quadrant, with the hypotenuse labeled –x. The horizontal leg is labeled –a and is on the negative x-axis. The vertical leg is labeled the square root of (x^2-a^2). To the right of the triangle is the equation sec(theta) = x/a, x<-a. There are also the equations sin(theta)= the negative square root of (x^2-a^2)/x, cos(theta) = a/x, and tan(theta) = the negative square root of (x^2-a^2)/a.
Use the appropriate reference triangle to express the trigonometric functions evaluated at θ in terms of x .

Finding the area of a region

Find the area of the region between the graph of f ( x ) = x 2 9 and the x -axis over the interval [ 3 , 5 ] .

First, sketch a rough graph of the region described in the problem, as shown in the following figure.

This figure is the graph of the function f(x) = the square root of (x^2-9). It is an increasing curve that starts on the x-axis at 3 and is in the first quadrant. Under the curve above the x-axis is a shaded region bounded to the right at x = 5.
Calculating the area of the shaded region requires evaluating an integral with a trigonometric substitution.

We can see that the area is A = 3 5 x 2 9 d x . To evaluate this definite integral, substitute x = 3 sec θ and d x = 3 sec θ tan θ d θ . We must also change the limits of integration. If x = 3 , then 3 = 3 sec θ and hence θ = 0 . If x = 5 , then θ = sec −1 ( 5 3 ) . After making these substitutions and simplifying, we have

Area = 3 5 x 2 9 d x = 0 sec −1 ( 5 / 3 ) 9 tan 2 θ sec θ d θ Use tan 2 θ = 1 sec 2 θ . = 0 sec −1 ( 5 / 3 ) 9 ( sec 2 θ 1 ) sec θ d θ Expand. = 0 sec −1 ( 5 / 3 ) 9 ( sec 3 θ sec θ ) d θ Evaluate the integral. = ( 9 2 ln | sec θ + tan θ | + 9 2 sec θ tan θ ) 9 ln | sec θ + tan θ | | 0 sec −1 ( 5 / 3 ) Simplify. = 9 2 sec θ tan θ 9 2 ln | sec θ + tan θ | | 0 sec −1 ( 5 / 3 ) Evaluate. Use sec ( sec −1 5 3 ) = 5 3 and tan ( sec −1 5 3 ) = 4 3 . = 9 2 · 5 3 · 4 3 9 2 ln | 5 3 + 4 3 | ( 9 2 · 1 · 0 9 2 ln | 1 + 0 | ) = 10 9 2 ln 3 .
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Evaluate d x x 2 4 . Assume that x > 2 .

ln | x 2 + x 2 4 2 | + C

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Key concepts

  • For integrals involving a 2 x 2 , use the substitution x = a sin θ and d x = a cos θ d θ .
  • For integrals involving a 2 + x 2 , use the substitution x = a tan θ and d x = a sec 2 θ d θ .
  • For integrals involving x 2 a 2 , substitute x = a sec θ and d x = a sec θ tan θ d θ .

Simplify the following expressions by writing each one using a single trigonometric function.

9 sec 2 θ 9

9 tan 2 θ

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a 2 + a 2 sinh 2 θ

a 2 cosh 2 θ

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Use the technique of completing the square to express each trinomial as the square of a binomial.

4 x 2 4 x + 1

4 ( x 1 2 ) 2

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x 2 2 x + 4

( x + 1 ) 2 + 5

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Integrate using the method of trigonometric substitution. Express the final answer in terms of the variable.

d x x 2 a 2

ln | x + a 2 + x 2 | + C

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d x 1 + 9 x 2

1 3 ln | 9 x 2 + 1 + 3 x | + C

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d x x 2 1 x 2

1 x 2 x + C

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x 2 + 9 d x

9 [ x x 2 + 9 18 + 1 2 l n | x 2 + 9 3 + x 3 | ] + C

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θ 3 d θ 9 θ 2 d θ

1 3 9 θ 2 ( 18 + θ 2 ) + C

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x 6 x 8 d x

( −1 + x 2 ) ( 2 + 3 x 2 ) x 6 x 8 15 x 3 + C

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d x ( 1 + x 2 ) 3 / 2

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d x ( x 2 9 ) 3 / 2

x 9 −9 + x 2 + C

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x 2 d x x 2 1

1 2 ( ln | x + x 2 1 | + x x 2 1 ) + C

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d x x 2 x 2 + 1

1 + x 2 x + C

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−1 1 ( 1 x 2 ) 3 / 2 d x

1 8 ( x ( 5 2 x 2 ) 1 x 2 + 3 arcsin x ) + C

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In the following exercises, use the substitutions x = sinh θ , cosh θ , or tanh θ . Express the final answers in terms of the variable x.

d x x 1 x 2

ln x ln | 1 + 1 x 2 | + C

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x 2 1 x 2 d x

−1 + x 2 x + ln | x + −1 + x 2 | + C

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1 + x 2 x 2 d x

1 + x 2 x + arcsinh x + C

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Use the technique of completing the square to evaluate the following integrals.

1 x 2 + 2 x + 1 d x

1 1 + x + C

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1 x 2 + 2 x + 8 d x

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1 x 2 + 10 x d x

2 −10 + x x ln | −10 + x + x | ( 10 x ) x + C

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1 x 2 + 4 x 12 d x

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Evaluate the integral without using calculus: −3 3 9 x 2 d x .

9 π 2 ; area of a semicircle with radius 3

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Find the area enclosed by the ellipse x 2 4 + y 2 9 = 1 .

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Evaluate the integral d x 1 x 2 using two different substitutions. First, let x = cos θ and evaluate using trigonometric substitution. Second, let x = sin θ and use trigonometric substitution. Are the answers the same?

arcsin ( x ) + C is the common answer.

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Evaluate the integral d x x x 2 1 using the substitution x = sec θ . Next, evaluate the same integral using the substitution x = csc θ . Show that the results are equivalent.

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Evaluate the integral x x 2 + 1 d x using the form 1 u d u . Next, evaluate the same integral using x = tan θ . Are the results the same?

1 2 ln ( 1 + x 2 ) + C is the result using either method.

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State the method of integration you would use to evaluate the integral x x 2 + 1 d x . Why did you choose this method?

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State the method of integration you would use to evaluate the integral x 2 x 2 1 d x . Why did you choose this method?

Use trigonometric substitution. Let x = sec ( θ ) .

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Evaluate −1 1 x d x x 2 + 1

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Find the length of the arc of the curve over the specified interval: y = ln x , [ 1 , 5 ] . Round the answer to three decimal places.

4.367

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Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x 2 , y = 0 , x = 0 , and x = 2 about the x -axis. (Round the answer to three decimal places).

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The region bounded by the graph of f ( x ) = 1 1 + x 2 and the x -axis between x = 0 and x = 1 is revolved about the x- axis. Find the volume of the solid that is generated.

π 2 8 + π 4

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Solve the initial-value problem for y as a function of x .

( x 2 + 36 ) d y d x = 1 , y ( 6 ) = 0

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( 64 x 2 ) d y d x = 1 , y ( 0 ) = 3

y = 1 16 ln | x + 8 x 8 | + 3

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Find the area bounded by y = 2 64 4 x 2 , x = 0 , y = 0 , and x = 2 .

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An oil storage tank can be described as the volume generated by revolving the area bounded by y = 16 64 + x 2 , x = 0 , y = 0 , x = 2 about the x -axis. Find the volume of the tank (in cubic meters).

24.6 m 3

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During each cycle, the velocity v (in feet per second) of a robotic welding device is given by v = 2 t 14 4 + t 2 , where t is time in seconds. Find the expression for the displacement s (in feet) as a function of t if s = 0 when t = 0 .

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Find the length of the curve y = 16 x 2 between x = 0 and x = 2 .

2 π 3

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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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