Question

Consider the following function. \int \frac{x^2}{\sqrt{x^2 - 5}} dx (a) Determine an appropriate trigonometric substitution. \circ Use x = \sqrt{5} \sin(\theta), where -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}, since the integrand contains the expression \sqrt{x^2 - 5}. \circ Use x = \sqrt{5} \tan(\theta), where -\frac{\pi}{2} < \theta < \frac{\pi}{2}, since the integrand contains the expression \sqrt{x^2 - 5}. \circ Use x = \sqrt{5} \sec(\theta), where 0 \le \theta < \frac{\pi}{2} or \pi \le \theta < \frac{3\pi}{2}, since the integrand contains the expression \sqrt{x^2 - 5}. (b) Apply the substitution to transform the integral into a trigonometric integral. Do not evaluate the integral. \int \frac{x^2}{\sqrt{x^2 - 5}} dx = \int (\qquad) d\theta

          Consider the following function.
\int \frac{x^2}{\sqrt{x^2 - 5}} dx
(a) Determine an appropriate trigonometric substitution.
\circ Use x = \sqrt{5} \sin(\theta), where -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}, since the integrand contains the expression \sqrt{x^2 - 5}.
\circ Use x = \sqrt{5} \tan(\theta), where -\frac{\pi}{2} < \theta < \frac{\pi}{2}, since the integrand contains the expression \sqrt{x^2 - 5}.
\circ Use x = \sqrt{5} \sec(\theta), where 0 \le \theta < \frac{\pi}{2} or \pi \le \theta < \frac{3\pi}{2}, since the integrand contains the expression \sqrt{x^2 - 5}.
(b) Apply the substitution to transform the integral into a trigonometric integral. Do not evaluate the integral.
\int \frac{x^2}{\sqrt{x^2 - 5}} dx = \int (\qquad) d\theta

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Consider the following function.
∫(x^2)/(√(x^2 - 5)) dx
(a) Determine an appropriate trigonometric substitution.
∘Use x = √(5) sin(θ), where -(π)/(2) ≤θ≤(π)/(2), since the integrand contains the expression √(x^2 - 5).
∘Use x = √(5) tan(θ), where -(π)/(2) < θ< (π)/(2), since the integrand contains the expression √(x^2 - 5).
∘Use x = √(5) sec(θ), where 0 ≤θ< (π)/(2) or π≤θ< (3π)/(2), since the integrand contains the expression √(x^2 - 5).
(b) Apply the substitution to transform the integral into a trigonometric integral. Do not evaluate the integral.
∫(x^2)/(√(x^2 - 5)) dx = ∫( ) dθ

Added by Amy O.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Sri K

07/31/2023

Step 1

Let x = sqrt(5)sec(theta), where 0 <= theta < pi/2 or pi <= theta < 3pi/2, since the integrand contains the expression sqrt(x^2 - 5). Show more…

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Consider the following function. int (x^(2))/(sqrt(x^(2)-5))dx (a) Determine an appropriate trigonometric substitution. Use x=sqrt(5)sin( heta ), where -(pi )/(2)<= heta <=(pi )/(2), since the integrand contains the expression sqrt(x^(2)-5). Use x=sqrt(5)tan( heta ), where -(pi )/(2)< heta <(pi )/(2), since the integrand contains the expression sqrt(x^(2)-5). Use x=sqrt(5)sec( heta ), where 0<= heta <(pi )/(2) or pi <= heta <(3pi )/(2), since the integrand contains the expression sqrt(x^(2)-5). (b) Apply the substitution to transform the integral into a trigonometric integral. Do not evaluate the integral. int (x^(2))/(sqrt(x^(2)-5))dx=int (◻)d heta Consider the following function. (a)Determine an appropriate trigonometric substitution 2 2 2 (b Apply the substitution to transform the integral into a trigonometric integral. Do not evaluate the integral. /V2-50x-J( 16

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00:01 Hello, let's have a look at the question.

00:02 So the question state that consider the following function.

00:06 So we have integration of x to the power 3 divided by under root 81 minus x square dx.

00:15 Now we have to determine in the a part the appropriate trigonometric substitution.

00:21 So firstly we have x is equals to 9 sine theta since the integrand contains the expression under root 81 minus x square or we can write this as under root 9 square minus x square.

00:38 So let us put the substitution here.

00:41 So we will get under root of 81 minus here we will have 9 sine theta whole square.

00:52 So this will be under root of 81 minus 81 sine square theta.

00:59 So further we can write taking in under root 81 common so we have 1 minus sine square theta.

01:07 So this will be 9 and under root of cos square theta as the value for 1 minus sine square theta is cos square theta.

01:17 So on solving we will get 9 cos theta.

01:21 So this is an appropriate substitution.

01:24 Now let us check for the second part that is x is equals to 9 tan theta.

01:32 So we have under root 9 square minus x square.

01:39 So we will get under root of 9 square minus 9 tan theta whole square.

01:48 So this will be under root of 9 square minus here we will get 9 square and tan square theta.

01:58 So taking 9 square common we have 9 square multiplied with 1 minus tan square theta.

02:07 So here this will be 9 under root of 1 minus tan square theta.

02:12 So here we see that this can't be solved further.

02:16 So this is not an appropriate substitution.

02:18 Now for the third part here we have that x is equals to 9 sec theta...

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