3.4E: Exercises for Section 7.3
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Simplify the expressions in exercises 1 - 5 by writing each one using a single trigonometric function.
1) 4−4sin2θ
2) 9sec2θ−9
- Answer
- 9sec2θ−9=9tan2θ
3) a2+a2tan2θ
4) a2+a2sinh2θ
- Answer
- a2+a2sinh2θ=a2cosh2θ
5) 16cosh2θ−16
Use the technique of completing the square to express each trinomial in exercises 6 - 8 as the square of a binomial.
6) 4x2−4x+1
- Answer
- 4(x−12)2
7) 2x2−8x+3
8) −x2−2x+4
- Answer
- −(x+1)2+5
In exercises 9 - 28, integrate using the method of trigonometric substitution. Express the final answer in terms of the original variable.
9) ∫dx√4−x2
10) ∫dx√x2−a2
- Answer
- ∫dx√x2−a2=ln∣x+√−a2+x2∣+C
11) ∫√4−x2dx
12) ∫dx√1+9x2
- Answer
- ∫dx√1+9x2=13ln∣√9x2+1+3x∣+C
13) ∫x2dx√1−x2
14) ∫dxx2√1−x2
- Answer
- ∫dxx2√1−x2=−√1−x2x+C
15) ∫dx(1+x2)2
16) ∫√x2+9dx
- Answer
- ∫√x2+9dx=9[x√x2+918+12ln|√x2+93+x3|]+C
17) ∫√x2−25xdx
18) ∫θ3√9−θ2dθ
- Answer
- ∫θ3dθ√9−θ2dθ=−13√9−θ2(18+θ2)+C
19) ∫dx√x6−x2
20) ∫√x6−x8dx
- Answer
- ∫√x6−x8dx=(−1+x2)(2+3x2)√x6−x815x3+C
21) ∫dx(1+x2)3/2
22) ∫dx(x2−9)3/2
- Answer
- ∫dx(x2−9)3/2=−x9√x2−9+C
23) ∫√1+x2xdx
24) ∫x2√x2−1dx
- Answer
- ∫x2√x2−1dx=12(ln∣x+√x2−1∣+x√x2−1)+C
25) ∫x2x2+4dx
26) ∫dxx2√x2+1
- Answer
- ∫dxx2√x2+1=−√1+x2x+C
27) ∫x2√1+x2dx
28) ∫1−1(1−x2)3/2dx
- Answer
- ∫1−1(1−x2)3/2dx=18(x(5−2x2)√1−x2+3arcsinx)+C
In exercises 29 - 34, use the substitutions x=sinhθ, coshθ, or tanhθ. Express the final answers in terms of the variable x.
29) ∫dx√x2−1
30) ∫dxx√1−x2
- Answer
- ∫dxx√1−x2=lnx−ln∣1+√1−x2∣+C
31) ∫√x2−1dx
32) ∫√x2−1x2dx
- Answer
- ∫√x2−1x2dx=−√−1+x2x+ln|x+√−1+x2|+C
33) ∫dx1−x2
34) ∫√1+x2x2dx
- Answer
- ∫√1+x2x2dx=−√1+x2x+arcsinhx+C
Use the technique of completing the square to evaluate the integrals in exercises 35 - 39.
35) ∫1x2−6xdx
36) ∫1x2+2x+1dx
- Answer
- ∫1x2+2x+1dx=−11+x+C
37) ∫1√−x2+2x+8dx
38) ∫1√−x2+10xdx
- Answer
- ∫1√−x2+10xdx=arcsin(x−55)+C
39) ∫1√x2+4x−12dx
40) Evaluate the integral without using calculus: ∫3−3√9−x2dx.
- Answer
- ∫3−3√9−x2dx=9π2; area of a semicircle with radius 3
41) Find the area enclosed by the ellipse x24+y29=1.
42) Evaluate the integral ∫dx√1−x2 using two different substitutions. First, let x=cosθ and evaluate using trigonometric substitution. Second, let x=sinθ and use trigonometric substitution. Are the answers the same?
- Answer
- ∫dx√1−x2=arcsin(x)+C is the common answer.
43) Evaluate the integral ∫dxx√x2−1 using the substitution x=secθ. Next, evaluate the same integral using the substitution x=cscθ. Show that the results are equivalent.
44) Evaluate the integral ∫xx2+1dx using the form ∫1udu. Next, evaluate the same integral using x=tanθ. Are the results the same?
- Answer
- ∫xx2+1dx=12ln(1+x2)+C is the result using either method.
45) State the method of integration you would use to evaluate the integral ∫x√x2+1dx. Why did you choose this method?
46) State the method of integration you would use to evaluate the integral ∫x2√x2−1dx. Why did you choose this method?
- Answer
- Use trigonometric substitution. Let x=sec(θ).
47) Evaluate ∫1−1xx2+1dx
48) Find the length of the arc of the curve over the specified interval: y=lnx,[1,5]. Round the answer to three decimal places.
- Answer
- s=4.367 units
49) Find the surface area of the solid generated by revolving the region bounded by the graphs of y=x2,y=0,x=0, and x=√2 about the x-axis. (Round the answer to three decimal places).
50) The region bounded by the graph of f(x)=11+x2 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.
- Answer
- V=(π28+π4)units3
In exercises 51 - 52, solve the initial-value problem for y as a function of x.
51) (x2+36)dydx=1,y(6)=0
52) (64−x2)dydx=1,y(0)=3
- Answer
- y=116ln|x+8x−8|+3
53) Find the area bounded by y=2√64−4x2,x=0,y=0, and x=2.
54) An oil storage tank can be described as the volume generated by revolving the area bounded by y=16√64+x2,x=0,y=0,x=2 about the x-axis. Find the volume of the tank (in cubic meters).
- Answer
- V=24.6 m3
55) During each cycle, the velocity v (in feet per second) of a robotic welding device is given by v=2t−144+t2, 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.
56) Find the length of the curve y=√16−x2 between x=0 and x=2.
- Answer
- s=2π3 units