Evaluate the following integral using trigonometric substitution. S dt 1²√36-1² OC. t-6 tan 0 Rewrite the given integral using this substitution. dt 1²√36-1² (Simplify your answers. Type exact answers.) Evaluate the integral. S = de dt 1² √36-1² (Type an exact answer.)

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 64E
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Evaluate the following integral using trigonometric substitution.
S
O C. t = 6 tan 0
Rewrite the given integral using this substitution.
dt
1²√36-1²
dt
SO de
1² √√36-1²
(Simplify your answers. Type exact answers.)
Evaluate the integral.
S
dt
+²√36-1²
(Type an exact answer.)
S
2
Transcribed Image Text:Evaluate the following integral using trigonometric substitution. S O C. t = 6 tan 0 Rewrite the given integral using this substitution. dt 1²√36-1² dt SO de 1² √√36-1² (Simplify your answers. Type exact answers.) Evaluate the integral. S dt +²√36-1² (Type an exact answer.) S 2
Evaluate the following integral using trigonometric substitution.
dt
1²2²√√36-1²
What substitution will be the most helpful for evaluating this integral?
OA. t=6 sec 0
O B. t=6 sin 0
OC. t = 6 tan 0
Rewrite the given integral using this substitution.
dt
= 10 do
1²2²√36-1²
(Simplify your answers. Type exact answers.)
s
Transcribed Image Text:Evaluate the following integral using trigonometric substitution. dt 1²2²√√36-1² What substitution will be the most helpful for evaluating this integral? OA. t=6 sec 0 O B. t=6 sin 0 OC. t = 6 tan 0 Rewrite the given integral using this substitution. dt = 10 do 1²2²√36-1² (Simplify your answers. Type exact answers.) s
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