7.6: Trigonometric Substitutions
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Integrals containing one of the terms
√a2+x2,√a2−x2, or √x2−a2
can often be integrated by a trigonometric substitution. The idea is to take x,a, and the square root as the three sides of a right triangle and use one of its acute angles as a new variable θ. The three kinds of trigonometric substitutions are shown in Figure 7.6.1. These figures do not have to be memorized. Just remember that the sides must be labeled so that
( opposite )2+( adjacent )2=( hypotenuse )2
These substitutions frequently give an integral of powers of trigonometric functions discussed in the preceding section.
x=atanθ
x=asinθ
x=asecθ
Figure 7:6.1
Find ∫(a2+x2)−3/2dx.
Solution
Let θ=arctan(x/a). Then from Figure 7.6.2,
x=atanθ,dx=asec2θdθ,√a2+x2=asecθ
So
∫(a2+x2)−3/2dx=∫(asecθ)−3asec2θdθ=1a2∫(secθ)−1dθ=1a2∫cosθdθ=1a2sinθ+C=1a2tanθsecθ+C=xa2√a2+x2+C
Find ∫√x2−a2dx.
Solution
Let θ=arcsec(x/a) (Figure 7.6.3), so
x=asecθ,dx=atanθsecθdθ,√x2−a2=atanθ
So
∫√x2−a2dx=∫atanθatanθsecθdθ=a2∫tan2θsecθdθ=a2∫(sec2θ−1)secθdθ=a2∫sec3θdθ−a2∫secθdθ=(12a2sec2θsinθ+12a2∫secθdθ)−a2∫secθdθ=12a2sec2θsinθ−12a2∫secθdθ=12x√x2−a2−12a2∫secθdθ
This is as far as we can go on this problem until we find out how to integrate ∫secθdθ in the next chapter.
∫1x2√a2−x2dx. Let θ=arcsin(x/a) (Figure 7.6.4). Then
Solution
x=asinθ,dx=acosθdθ,√a2−x2=acosθ∫1x2√a2−x2dx=∫1a2sin2θacosθacosθdθ=∫1a2sin2θdθ=1a2∫csc2θdθ=−1a2cotθ+C=−1a2√a2−x2x+C
∫√x2−a2xdx. Put θ=arcsec(x/a) (Figure 7.6.5). Then
Solution
x=asecθ,dx=atanθsecθdθ,√x2−a2=atanθ∫√x2−a2xdx=∫atanθasecθatanθsecθdθ=a∫tan2θdθ⋅=a∫sec2θdθ−a∫dθ=atanθ−aθ+C=√x2−a2−aarcsec(x/a)+C
Figure 7.6 .4
To keep track of a trigonometric substitution, it is a good idea to actually draw the triangle and label the sides.
The basic integrals:
Solution
(a)
∫1√1−x2dx=arcsinx+C
(b)
∫dx1+x2dx=arctanx+C
(c)
∫dxx√x2−1=arcsecx+C,x>1
can be evaluated very easily by a trigonometric substitution.
(a)
∫1√1−x2dx
Let θ=arcsinx (Figure 7.6.6). Then x=sinθ,dx=cosθdθ,√1−x2= cosθ.
∫1√1−x2dx=∫cosθdθcosθ=∫dθ=θ+C∫1√1−x2dx=arcsinx+C
(b)
∫dx1+x2
Let θ=arctanx (Figure 7.6.7). Then x=tanθ,dx=sec2θdθ,√1+x2= secθ.
∫dx1+x2=∫sec2θsec2θdθ=∫dθ=θ+C∫dx1+x2=arctanx+C
(c)
∫dxx√x2−1,x>1
Let θ=arcsecx (Figure 7.6.8). Then x=secθ,dx=tanθsecθdθ, √x2−1=tanθ.
∫dxx√x2−1=∫tanθsecθsecθtanθdθ=∫dθ=θ+C∫dxx√x2+1=arcsecx+C,x>1
It is therefore more important to remember the method of trigonometric substitution than to remember the integration formulas (a), (b), (c).
PROBLEMS FOR SECTION 7.6
Draw the appropriate triangle and evaluate using trigonometric substitutions
1∫dx√1−4x2
2∫√a2−x2dx
3∫x3dx√9+¯x2
4∫√x2−1xdx
5∫(4−x2)−3/2dx
6∫(1−3x2)3/2dx
7∫sinθdθ√2−cos2θ
8∫dx√x√1−x
9∫dxx2(1+x2)
10∫x4dx9+x2
11∫x2dx√4−x2
12∫√9−4x2x4dx
13∫x2√1−x2dx
14 ∫√x√1−xdx
15
∫√4x−x2dx
16∫√x√1−xdx
17∫x3√4x2−1dx
18∫−dxx3√x2−3
19∫x3√a2−x2dx
20∫x3√1+a2x2dx
21∫√x4−1xdx
22∫x√1−x4dx
23 ∫x3(a2+x2)3/2dx
24
∫20√4−x2dx
25 ∫1−1x2√1−x2dx
∫40dx(9+x2)3/2
27∫∞0dx(9+x2)3/2
∫42√x2−2xdx
29∫x2√x2−2xdx
30∫x0x3√1+x2dx
31∫xarcsinxdx
32∫xarccosxdx
33∫x2arcsinxdx
34∫x3arctanxdx
35 ∫x−3arcsinxdx
36∫x−3arctanxdx
37 Find the surface area generated by rotating the ellipse x2+4y2=1 about the x-axis.