25Evaluate the integral:dxx2 + 64(A) Which trig substitution is correct for this integral?O x =8 tan(0)64 tan(0)O x =8 sin(0)O x =64 sin(0)х — 25 sec(Ө)64 sec(0)(B) Which integral do you obtain after substituting for x and simplifying?Note: to enter 0, type the word theta.do(C) What is the value of the above integral in terms of 0?+ C(D) What is the value of the original integral in terms of x?+ C

Given that : The integral is ∫25x2 x2 + 64 dx .

Answered: O x = 8 tan(0)

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter7: Analytic Trigonometry

Section7.2: Trigonometric Equations

Problem 75E

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Related questions

Question

25
Evaluate the integral:
dx
x2 + 64
(A) Which trig substitution is correct for this integral?
O x =
8 tan(0)
64 tan(0)
O x =
8 sin(0)
O x =
64 sin(0)
х — 25 sec(Ө)
64 sec(0)
(B) Which integral do you obtain after substituting for x and simplifying?
Note: to enter 0, type the word theta.
do
(C) What is the value of the above integral in terms of 0?
+ C
(D) What is the value of the original integral in terms of x?
+ C

Expert Solution

Step 1

Given that :

The integral is $\int \frac{25}{x^{2} \sqrt{x^{2} + 64}} d x$ .

Step 2

By using formulae,

tan² x + 1 = sec² x.

Step 3

To find the integral $\int \frac{25}{x^{2} \sqrt{x^{2} + 64}} d x$ :

By using substitution.

Substitute x = 8 tan $(θ)$ , then dx = $s e c^{2} (θ) d θ$ .

Then,

$\int \frac{25}{x^{2} \sqrt{x^{2} + 64}} d x = 25 \int \frac{1}{x^{2} \sqrt{x^{2} + 64}} d x = 25 \int \frac{8 s e c^{2} (θ) d θ}{64 \tan^{2} (θ) \sqrt{64 \tan^{2} (θ) + 64}} = 25 \int \frac{8 s e c^{2} (θ) d θ}{64 \tan^{2} (θ) \sqrt{64 (\tan^{2} (θ) + 1)}}$

Step 4

$= 25 \int \frac{8 s e c^{2} (θ) d θ}{64 \tan^{2} (θ) \sqrt{64 s e c^{2} (θ)}} = 25 \int \frac{8 s e c^{2} (θ)}{64 \tan^{2} (θ) . 8 s e c (θ)} d θ = 25 \int \frac{s e c (θ)}{64 \tan^{2} (θ)} d θ = \frac{25}{64} \int \frac{s e c (θ)}{\tan^{2} (θ)} d θ = \frac{25}{64} \int \frac{\frac{1}{\cos (θ)}}{\frac{\sin^{2} (θ)}{\cos^{2} (θ)}} d θ = \frac{25}{64} \int \frac{\cos (θ)}{\sin^{2} (θ)} d θ$

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