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Trigonometric Integrands and Trigonometric Substitutions Worksheet

Trigonometric Integrands and Trigonometric Substitutions Worksheet
  • Page 1
 1.  
Find 20a2 -x2 dx.
a.
Sin-1xa + C
b.
20 Sin-1xa + C
c.
Sin-1ax + C
d.
20 Sin-1ax + C


Solution:

Let x = a sin θ θ = Sin-1(xa) , dx = a cos θ dθ

20a2 -x2 dx = 20a² - (a sin θ)²  a cos θ dθ

= 20acosθa2(1 -sin2θ)d θ

= 20acosθacosθd θ

= 20 d θ

= 20(θ) + C

= 20Sin-1(xa) + C
[Substitute θ = Sin-1(xa).]


Correct answer : (2)
 2.  
Find 31x2 -a2dx.
a.
31 Sin-1(xa) + C
b.
31 sinh-1(xa) + C
c.
31 cosh-1(ax) + C
d.
31 cosh-1(xa) + C


Solution:

Let x = a cosh θ θ = cosh-1(xa), dx = a sinh θ d θ

31x2 -a2dx = 31a2cosh2 θ -a2asinhθdθ

= 31asinhθasinhθd θ

= 31 d θ

= 31(θ) + C

= 31cosh-1(xa) + C
[Substitute θ = cosh-1(xa).]


Correct answer : (4)
 3.  
Find 36a2+x2dx.
a.
36aTan-1(xa) + C
b.
36 aTan-1(xa) + C
c.
36aTan-1(ax) + C
d.
36 Tan-1(1x) + C


Solution:

Let x = a tan θ θ = Tan-1(xa) and dx = a sec2 θ d θ

36a2+x2dx = 36a2+a2tan2θasec2 θd θ

= 36asec2 θa2sec2 θd θ

= 36a d θ

= 36a (θ) + C

= 36aTan-1(xa) + C
[Substitute θ = Tan-1(xa).]


Correct answer : (2)
 4.  
Evaluate 28(x2+a2)dx.
a.
28 aTan-1(xa) + C
b.
28 tanh-1(xa) + C
c.
28 sinh-1(xa) + C
d.
28 cosh-1(xa) + C


Solution:

28(x2+a2)dx = 28(asinhθ)2+a2acosh θ d θ
[x = a sinh θ dx = a cosh θ d θ.]

= 28acoshθacoshθd θ

= 28d θ

= 28θ + C

= 28sinh-1xa + C
[Substitute θ = sinh-1xa.]


Correct answer : (3)
 5.  
Evaluate 114+x2dx.
a.
11 2Tan-1(x2 ) + C
b.
Cot-1(x2 ) + C
c.
Tan-1(x2 ) + C
d.
2 Tan-1(x2 ) + C


Solution:

Let x = 2tan θ θ = Tan-1x2 and dx = 2 sec2 θ d θ

114+x2dx = 114+(2 tan θ)2(2sec2θ)d θ

= 112sec2θ4sec2 θd θ

= 11 / 2 d θ

= 11 / 2θ + C

= 11 / 2Tan-1(x2) + C
[Substitute θ = Tan-1(x2).]


Correct answer : (1)
 6.  
Simplify 12xTan-1(x2)dx.
a.
6 x2 tan x + ln | 1 + x | + C
b.
1 2(x2 tan(x2)) + ln | 1+x4 | + C
c.
6 x2 tan x + ln | 1 + x | + C
d.
6 (x2Tan-1(x)2 - ln |1+x4|) + C


Solution:

Let x2 = tan θ θ = Tan-1(x2) and x dx = 1 / 2sec2 θ d θ

12xTan-1(x2)dx = 12Tan-1(tan θ) 1 / 2sec2 θ d θ

= 6 θ sec2 θ d θ

= 6 (θ tanθ - tan θd θ)
[Use integration by parts.]

= 6 (θ tanθ + ln |cos θ |) + C
[Use tan x dx = - ln |cos x| + C.]

= 6 (x2 Tan-1(x2) + ln |11+x4|) + C
[ Substitute cos θ = 11+x4.]

= 6 (x2 Tan-1(x)2 - ln |1+x4|) + C
[Substitute θ = Tan-1(x2).]


Correct answer : (4)
 7.  
Evaluate Tan-11 - 4x1 + 4xdx.
a.
1 8(4xCos-14x + 1 - 16x2) + C
b.
1 8(4xCos-14x - 1 - 16x2) + C
c.
(4xCos-14x - 1 - 16x2) + C
d.
1 8(1 - 16x2) + C


Solution:

Tan-11 - 4x1 + 4xdx

Let 4x = cos θ θ = Cos-14x and dx = - 1 / 4sin θ d θ

= Tan-11 - cos θ1 + cos θ(- 1 / 4sin θ)d θ

= - 1 / 4sin θ Tan-12sin2θ22cos2θ2d θ
[Use half angle formulae.]

= - 1 / 4sin θ Tan-1(tanθ2)d θ

= - 1 / 8θ sin θd θ

= - 1 / 8(- θcosθ + cos θd θ)
[Use integration by parts.]

= 1 / 8(θcos θ - sin θ)

= 1 / 8(4xCos-14x - 1 - 16x2) + C
[Substitute sin θ = 1 - 16x2, cos θ = 4x, θ = Cos-14x.]


Correct answer : (2)
 8.  
Find cos 3xsin 3x(1 + sin 3x)dx.
a.
3 ln |sin 3x1 + sin 3x | + C
b.
ln |sin 3x1 + sin 3x | + C
c.
1 3ln |sin 3x1 + sin 3x | + C
d.
1 3ln |sinx1 + sinx | + C


Solution:

Let sin 3x = t 3x = Sin-1t and cos 3x dx = 1 / 3dt

cos 3xsin 3x(1 + sin 3x)dx = 1 / 3dtt(t + 1)

= 1 / 3(1t - 11 + t)dt
[Use the method of partial fractions.]

= 1 / 3(1tdt - 11 + tdt)

= 1 / 3(ln t - ln|1 + t|) + C

= 1 / 3ln |t1 + t| + C

= 1 / 3ln |sin 3x1 + sin 3x | + C
[Substitute t = sin 3x.]


Correct answer : (3)
 9.  
Evaluate 21cos 18xcos 9x-sin 9xdx.
a.
9(sin 9x + cos 9x) + C
b.
sin 9x - cos 9x + C
c.
7 3(sin 9x - cos 9x) + C
d.
9(sin 9x - cos 9x) + C


Solution:

21cos 18xcos 9x-sin 9xdx = 21(cos2 9x-sin2 9x)cos 9x-sin 9xdx
[Use cos 2x = cos2 x - sin2 x.]

= 21(cos 9x+sin 9x)(cos 9x-sin 9x)cos 9x-sin 9xdx
[Use a2 - b2 = (a - b)(a + b).]

= 21(cos 9x + sin 9x) dx

= 21cos 9x dx + 21sin 9x dx

= 7 / 3(sin 9x - cos 9x) + C


Correct answer : (3)
 10.  
Find Tan-1(6x1-9x2)dx.
a.
2 3(ln |1 + 9x2|) + C
b.
2 3(3xTan-13x + ln |1 + 9x2|) + C
c.
2 3(3xTan-13x - ln |1 + 9x2|) + C
d.
(3xTan-13x - ln |1 + 9x2|) + C


Solution:

Let 3x = tan θ θ = Tan-13x and dx = 1 / 3sec2 θ dθ

Tan-1(6x1 - 9x2)dx = Tan-1(2tanθ1-tan2θ) 1 / 3sec2θd θ

= 1 / 3Tan-1(tan2θ)sec²θdθ

= 1 / 32θsec2 θ d θ

= 2 / 3θsec2 θd θ

= 2 / 3(θtan θ - (tan θ)d θ)
[Use integration by parts.]

= 2 / 3(θtanθ + ln |cos θ|) + C

= 2 / 3(3xTan-1(3x) + ln |11+9x2|) + C
[Substitute 3x = tan θ, θ = Tan-13x and cos θ = 11+9x2.]

= 2 / 3(3xTan-13x - ln |1 + 9x2|) + C


Correct answer : (3)

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