Section 7.3 Trigonometric Substitution

Section 7.3 Trigonometric Substitution

30. Evaluate the integral. R1 0

√

x − x²dx Solution:

SECTION 7.3 TRIGONOMETRIC SUBSTITUTION ¤ 647 23.

 

√²+ 2 + 5=

 

( + 1)²+ 4 =

 2 sec² 

√4 tan² + 4

 + 1 = 2 tan ,

 = 2 sec² 



=

 2 sec² 

2 sec  =



sec   = ln |sec  + tan | + ¹

= ln





√²+ 2 + 5

2 + + 1

2



 + ¹,

or ln√²+ 2 + 5 +  + 1 + , where  = 1− ln 2.

24.  1 0

 − ² =

 1 0

1 4 −

²−  +¹4

 =

 1 0

1 4 −

 −¹2

2



=

 2

−2

1

4− ¹4sin² ¹₂cos  

− ¹2= ¹₂sin ,

 = ¹₂cos  



= 2

 2 0

1

2cos ¹₂cos   = ¹₂

 2 0

cos²  =¹₂

 2 0

1

2(1 + cos 2) 

= ¹₄

 +¹₂sin 22 0 = ¹₄_

2

= ^₈

25. 

²√

3 + 2 − ² =

²

4 − (²+ 2 + 1)  =

²

2²− ( − 1)²

=

(1 + 2 sin )²√

4 cos² 2 cos  

− 1 = 2 sin ,

 = 2 cos  



=

(1 + 4 sin  + 4 sin²) 4 cos² 

= 4

(cos² + 4 sin  cos² + 4 sin² cos²) 

= 4 1

2(1 + cos 2)  + 4

4 sin  cos²  + 4

(2 sin  cos )²

= 2

(1 + cos 2)  + 16

sin  cos²  + 4

sin²2 

= 2

 +¹₂sin 2 + 16

−¹3cos³ + 4 ₁

2(1 − cos 4) 

= 2 + sin 2 −¹⁶3 cos³ + 2

 −¹4sin 4 + 

= 4 −¹2sin 4 + sin 2 −¹⁶3 cos³ + 

= 4 −¹2(2 sin 2 cos 2) + sin 2 −¹⁶3 cos³ + 

= 4 + sin 2(1 − cos 2) −¹⁶3 cos³ + 

= 4 + (2 sin  cos )(2 sin²) −¹⁶3 cos³ + 

= 4 + 4 sin³ cos  −¹⁶3 cos³ + 

= 4 sin⁻¹

 − 1 2

 + 4

 − 1 2

3√

3 + 2 − ²

2 −16

3

(3 + 2 − ²)^32

2³ + 

= 4 sin⁻¹

 − 1 2

 +1

4( − 1)³√

3 + 2 − ²−2

3(3 + 2 − ²)^32+  26. 3 + 4 − 4² = −(4²− 4 + 1) + 4 = 2²− (2 − 1)².

Let 2 − 1 = 2 sin , so 2  = 2 cos   and√

3 + 4 − 4²= 2 cos .

Then

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44. Find the volume of the solid obtained by rotating about the line x = 1 the region under the curve y = x√ 1 − x², 0 ≤ x ≤ 1.

Solution:

SECTION 7.3 TRIGONOMETRIC SUBSTITUTION ¤ 651

38. Use shells about  = 1:

 =1

0 2(1 − ) √

1 − ²

= 21 0 √

1 − ² − 21 0 ²√

1 − ² = 21− 2² For 1, let  = 1 − ², so  = −2 , and

1=0 1

√

−¹2

= ¹₂1

0 ^12 = ¹₂₂

3^321 0= ¹₂2

3

= ¹₃.

For 2, let  = sin , so  = cos  , and

2=2

0 sin²√

cos² cos   =2

0 sin² cos²  =2 0

1

4(2 sin  cos )²

= ¹₄2

0 sin²2  = ¹₄2 0

1

2(1 − cos 2)  = ¹8

 −¹2sin 22 0 = ¹₈

2

= ₁₆^

Thus,  = 21 3

− 2 16

= ²₃ −¹8².

39. (a) Let  =  sin ,  =  cos  ,  = 0 ⇒  = 0 and  =  ⇒

 = sin⁻¹(). Then

  0

²− ² =

 sin⁻¹() 0

 cos  ( cos  ) = ²

 sin⁻¹() 0

cos² 

= ² 2

 sin⁻¹() 0

(1 + cos 2)  = ² 2

 +¹₂sin 2sin⁻¹()

0 = ²

2

 + sin  cos sin⁻¹() 0

= ² 2



sin⁻¹ 



 +

 ·

√²− ²





− 0



= ¹₂²sin⁻¹() +¹₂√

²− ²

(b) The integral 0

√²− ²represents the area under the curve  =√

²− ²between the vertical lines  = 0 and  = .

The figure shows that this area consists of a triangular region and a sector of the circle ²+ ²= ². The triangular region has base  and height√

²− ², so its area is¹₂√

²− ². The sector has area ¹₂² = ¹₂²sin⁻¹().

40. The curves intersect when ²+₁

2²2

= 8 ⇔ ²+¹₄⁴= 8 ⇔ ⁴+ 4²− 32 = 0 ⇔ (²+ 8)(²− 4) = 0 ⇔  = ±2. The area inside the circle and above the parabola is given by

1=2

−2

√8 − ²− ¹2²

 = 22 0

√8 − ² − 22 0

1 2²

= 2

1

2(8) sin⁻¹

√2 8

+¹₂(2)√

8 − 2²−¹2

₁

3³2 0

 [by Exercise 39]

= 8 sin⁻¹

√1 2

+ 2√

4 −⁸3 = 8_

4

+ 4 − ⁸3 = 2 + ⁴₃

Since the area of the disk is √

8²= 8, the area inside the circle and below the parabola ia 2= 8 −

2 +⁴₃

= 6 − ⁴3.

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45. (a) Use trigonometric substitution to verify that Z x

0

pa²− t²dt = 1

2a²sin⁻¹x a

 +1

2xp a²− x²

(b) Use the figure to give trigonometric interpretations of both terms on the right side of the equation in part (a).

492 Chapter 7 Techniques of Integration

32. Evaluate

y

_sx²₁^x2_a²_d^{3y2 dx}

(a) by trigonometric substitution.

(b) by the hyperbolic substitution x − a sinh t.

33. Find the average value of fsxd −sx²21yx, 1 < x < 7.

34. Find the area of the region bounded by the hyperbola 9x²24y²− 36 and the line x − 3.

35. Prove the formula A −¹2r² for the area of a sector of a circle with radius r and central angle . [Hint: Assume 0 ,  , y2 and place the center of the circle at the origin so it has the equation x²1y²− r². Then A is the sum of the area of the triangle POQ and the area of the region PQR in the figure.]

O x

y

R

¨ Q

P

36. Evaluate the integral

y

_x⁴_sx^dx2₂₂

Graph the integrand and its indefinite integral on the same screen and check that your answer is reasonable.

37. Find the volume of the solid obtained by rotating about the x-axis the region enclosed by the curves y − 9ysx²19d, y − 0, x − 0, and x − 3.

38. Find the volume of the solid obtained by rotating about the line x − 1 the region under the curve y − xs1 2 x² , 0 < x < 1.

39. (a) Use trigonometric substitution to verify that

y

0^xsa²2t² dt −¹2a² sin²¹sxyad 1¹₂xsa²2x²

;

(b) Use the figure to give trigonometric interpretations of both terms on the right side of the equation in part (a).

¨ ¨

y=œ„„„„„a@-t@

0 t y a

x

40. The parabola y −¹₂x² divides the disk x²1y²<8 into two parts. Find the areas of both parts.

41. A torus is generated by rotating the circle x²1sy 2 Rd²− r² about the x-axis. Find the volume enclosed by the torus.

42. A charged rod of length L produces an electric field at point Psa, bd given by

EsPd −

y

2a^L2a

b

4«0sx²1b²d^3y2 dx

where  is the charge density per unit length on the rod and

«0 is the free space permittivity (see the figure). Evaluate the integral to determine an expression for the electric field EsPd.

0 x

y

L P (a, b)

43. Find the area of the crescent-shaped region (called a lune) bounded by arcs of circles with radii r and R. (See the figure.)

R r

44. A water storage tank has the shape of a cylinder with diam- eter 10 ft. It is mounted so that the circular cross-sections are vertical. If the depth of the water is 7 ft, what per cent age of the total capacity is being used?

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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Solution:

1

SECTION 7.3 TRIGONOMETRIC SUBSTITUTION ¤ 651

38. Use shells about  = 1:

 =1

0 2(1 − ) √

1 − ²

= 21 0 √

1 − ² − 21 0 ²√

1 − ² = 21− 2² For 1, let  = 1 − ², so  = −2 , and

1=0 1

√

−¹2

= ¹₂1

0 ^12 = ¹₂

2 3^321

0= ¹₂₂

3

= ¹₃.

For 2, let  = sin , so  = cos  , and

2=2

0 sin²√

cos² cos   =2

0 sin² cos²  =2 0

1

4(2 sin  cos )²

= ¹₄2

0 sin²2  = ¹₄2 0

1

2(1 − cos 2)  = ¹8

 −¹2sin 22 0 = ¹₈

2

= ₁₆^

Thus,  = 21 3

− 2 16

= ²₃ −¹8².

39. (a) Let  =  sin ,  =  cos  ,  = 0 ⇒  = 0 and  =  ⇒

 = sin⁻¹(). Then

  0

²− ² =

 sin⁻¹() 0

 cos  ( cos  ) = ²

 sin⁻¹() 0

cos² 

= ² 2

 sin⁻¹() 0

(1 + cos 2)  = ² 2

 +¹₂sin 2sin⁻¹()

0 = ²

2

 + sin  cos sin⁻¹() 0

= ² 2



sin⁻¹ 



 +

 ·

√²− ²





− 0



= ¹₂²sin⁻¹() + ¹₂√

²− ²

(b) The integral 0

√²− ²represents the area under the curve  =√

²− ²between the vertical lines  = 0 and  = .

The figure shows that this area consists of a triangular region and a sector of the circle ²+ ²= ². The triangular region has base  and height√

²− ², so its area is¹₂√

²− ². The sector has area ¹₂² = ¹₂²sin⁻¹().

40. The curves intersect when ²+₁

2²2

= 8 ⇔ ²+ ¹₄⁴= 8 ⇔ ⁴+ 4²− 32 = 0 ⇔ (²+ 8)(²− 4) = 0 ⇔  = ±2. The area inside the circle and above the parabola is given by

1=2

−2

√8 − ²−¹2²

 = 22 0

√8 − ² − 22 0

1 2²

= 2

1

2(8) sin⁻¹

√2 8



+¹₂(2)√

8 − 2²− ¹2

1 3³2

0

 [by Exercise 39]

= 8 sin⁻¹

√1 2

+ 2√

4 −⁸3 = 8_

4

+ 4 − ⁸3 = 2 +⁴₃

Since the area of the disk is √

8²= 8, the area inside the circle and below the parabola ia 2= 8 −

2 + ⁴₃

= 6 −⁴3.

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47. A torus is generated by rotating the circle x²+ (y − R)²= r² about the x-axis. Find the volume enclosed by the torus.

Solution:

652 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION

41. We use cylindrical shells and assume that   . ² = ²− ( − )² ⇒  = ±

²− ( − )², so () = 2

²− ( − )²and

 =+

− 2 · 2

²− ( − )² =

−4( + )√

²− ² [where  =  − ]

= 4

−√

²− ² + 4

−

√²− ²

where  =  sin  ,  =  cos  

in the second integral



= 4

−¹3(²− ²)^32

−+ 42

−2²cos²  = −^43 (0 − 0) + 4²2

−2cos² 

= 2²2

−2(1 + cos 2)  = 2²

 +¹₂sin 22

−2 = 2²² Another method: Use washers instead of shells, so  = 8

0

²− ²as in Exercise 6.2.63(a), but evaluate the

integral using  =  sin .

42. Let  =  tan , so that  =  sec² and√

²+ ²=  sec .

( ) =

 −

−



40(²+ ²)^32  = 

40

 ₂

₁

1

( sec )³  sec² 

= 

40

 ₂

₁

1

sec  =  40

 ₂

₁

cos   =  40

sin ₂

₁

= 

40

 

√²+ ²

−

−

= 

40

  − 

( − )²+ ² + 

√²+ ²



43. Let the equation of the large circle be ²+ ²= ². Then the equation of the small circle is ²+ ( − )²= ², where  =√

²− ²is the distance between the centers of the circles. The desired area is

 =

−

 +√

²− ²

−√

²− ²



= 2 0

 +√

²− ²−√

²− ²



= 2

0   + 2 0

√²− ² − 2 0

√²− ²

The first integral is just 2 = 2√

²− ². The second integral represents the area of a quarter-circle of radius , so its value is¹₄². To evaluate the other integral, note that

 √²− ² =

²cos²  [ =  sin ,  =  cos  ] =₁

2²

(1 + cos 2) 

= ¹₂²

 + ¹₂sin 2

+  = ¹₂²( + sin  cos ) + 

= ²

2 arcsin 



+² 2

 



 √²− ²

 +  = ²

2 arcsin 



+ 2

√²− ²+ 

Thus, the desired area is

 = 2√

²− ²+ 21 4²

−

²arcsin() + √

²− ²^

0

= 2√

²− ²+¹₂²−

²arcsin() + √

²− ²

= √

²− ²+^₂²− ²arcsin()

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