We have studied the use of trigonometric substitution in finding the antiderivat

Question

We have studied the use of trigonometric substitution in finding the antiderivatives of certain special types of functions. The success of these substitutions depends on certain trigonometric identities one the integrals of special combinations of trigonometric functions. In this problem, you will learn that hyperbolic functions may be in place of the trigonometric functions in the substitution method for finding antiderivatives. a) Review the hyperbolic functions in Section 3.11 of our text book (Early Transcendentals, 8th Edition, James Stewart) Then, use the appropriate hyperbolic Function substitutions and identities to find the following integrals. Clearly identify what functions and identies that you use in your work. b) Use appropriate trigonometric substitutions to find the same integrals. i) integral (9x^2, 4)^3/4 dx ii) integral 1/k^2 Squareroot 16 -x^2 dx

Explanation / Answer

Solution:3b(i)

use substitution x = a tan(u)
Then x² = a² tan²(u), and 9x² + 4 = 9a² tan²u + 4

Since we want to end up with tan²u + 1 in denominator, we need 9a² = 4, a = 2/3

x = (2/3) tan(u)
dx = (2/3) sec²u du

1/(9x²+4)^3/2 dx
= 1/(9((2/3) tan(u))²+4)^3/2 * (2/3)sec²u du
= (2/3) sec²(u) / (4tan²u + 4)^3/2 du
= (2/3) sec²u / (4(tan²u+1))^3/2 du
= (2/3)sec²u / (4sec²u)^3/2 du
= (2/3)sec²u / (8sec³u) du
= 1/12 1 / sec u du
= 1/12 cos u du
= 1/12 sin u + C

x = (2/3) tan(u)
tan(u) = 3x/2
sin(u)/cos(u) = 3x/2
sin u = 1 /(1 + (2/3x)^2) = 3x /(9x²+2²) = 3x/(9x²+4)

1/(9x²+4)^3/2 dx = 1/4 sin u + C
1/(9x²+4)^3/2 dx = 3x/(4(9x²+4)) + C

Solution: 3b(ii)

1/(x2 sqrt(16 - x2)) dx

Trig substitution

x = 4sint, dx= 4cost dt

= [1/ [16sin2t (16 - 16sin2t)]] * 4cost dt
= [1/ [16sin2t 16(1-sin2t)]] * 4cost dt
= [1/ [16sin2t 16(cos2t)]] * 4cost dt
= [1/ [16sin2t 4cost]] * 4cost dt
= [1/(16sin2t] dt
= (csc2t / 16) dt

= - cot(t) / 16 + C

substitue back t = arcsin(x/4)

= - cot(arcsin(x/4)) / 16 + C

=[-(16-x²) / 16x] + C

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