4. 16x – 2x2 – 23 dx -

Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter8: Further Techniques And Applications Of Integration
Section8.2: Integration By Parts
Problem 32E
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Trigonometric Substitution
There are certain integrals that contain no trigonometry that can be solved with the aid of
trigonometry. Consider integrands involving va? -x, a? +x, or vx - a (where a is constant).
These can be evaluated by making the following substitutions.
Va? + x
x= asin 6
x= atan 0
X= asec 0
dx = acos ede
dx = a sec*o de
dx = asec Otan ede
For each integral, figure out which trigonometric substitution should be used. The goal is to rewrite
the integrand so that there is no radical. After integrating, draw and label a right triangle (as shown
below). This will allow you to rewrite your integrated expression back in terms of x. For this last part
of the problem, you may need one or more
trigonometric identities.
For example, the triangle that corresponds with
exercise 1 looks like this. If we let x = 2sin 0, it follows
that sin 6 = x/2. By expressing the side length opposite
the given angle as x and the hypotenuse as 2, the side
length adjacent the given angle works out conveniently
to 4 -x, which is the integrand.
!3!
Transcribed Image Text:Trigonometric Substitution There are certain integrals that contain no trigonometry that can be solved with the aid of trigonometry. Consider integrands involving va? -x, a? +x, or vx - a (where a is constant). These can be evaluated by making the following substitutions. Va? + x x= asin 6 x= atan 0 X= asec 0 dx = acos ede dx = a sec*o de dx = asec Otan ede For each integral, figure out which trigonometric substitution should be used. The goal is to rewrite the integrand so that there is no radical. After integrating, draw and label a right triangle (as shown below). This will allow you to rewrite your integrated expression back in terms of x. For this last part of the problem, you may need one or more trigonometric identities. For example, the triangle that corresponds with exercise 1 looks like this. If we let x = 2sin 0, it follows that sin 6 = x/2. By expressing the side length opposite the given angle as x and the hypotenuse as 2, the side length adjacent the given angle works out conveniently to 4 -x, which is the integrand. !3!
4.
V16x- 2x2-23 dx
Transcribed Image Text:4. V16x- 2x2-23 dx
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