(i) Evaluate integral [ x^3 / (x^2 + 4)^2 ] using trigonometric substitution.
(ii) Evaluate integral [ x^3 / (x^2 + 4)^2 ] using regular substitution.
(iii) Use a right triangle to check that indeed both answers you obtained in
parts (i) and (ii) are the same.
I would really appreciate some help with this question. I've been stuck on it for 2 hours.
https://www.freemathhelp.com/forum/threads/49663-trig-substitution-int-x-3-sqrt-x-2-4-dx
Sure, I'd be happy to help you with this question!
(i) To evaluate the integral [ x^3 / (x^2 + 4)^2 ] using trigonometric substitution, we can make the substitution x = 2tanθ.
First, let's find the derivative of x with respect to θ using the chain rule:
dx/dθ = 2sec^2(θ)
Next, we can replace x and dx in the integral with expressions involving θ:
x = 2tanθ
dx = 2sec^2(θ)dθ
Now let's substitute these expressions into the original integral:
∫ [ x^3 / (x^2 + 4)^2 ] dx = ∫ [ (2tanθ)^3 / ((2tanθ)^2 + 4)^2 ] (2sec^2(θ)) dθ
= 8∫ [ tan^3(θ) / (4tan^2(θ) + 4)^2 ] sec^2(θ) dθ
Next, we can simplify the integrand using trigonometric identities:
tan^3(θ) = (sin^3(θ) / cos^3(θ)) = (sinθ / cos^2(θ)) = secθtan(θ)
sec^2(θ) = 1 + tan^2(θ)
Now our integral becomes:
8∫ [ secθtan(θ) / (4tan^2(θ) + 4)^2 ] (1 + tan^2(θ)) dθ
= 8∫ [ secθtan(θ) / 4(1 + tan^2(θ) + 4)^2 ] (1 + tan^2(θ)) dθ
= 2∫ [ secθtan(θ) / (1 + tan^2(θ) + 4)^2 ] dθ
At this point, we can make another substitution u = tan(θ):
du = sec^2(θ) dθ
Now, our integral becomes:
2∫ [ 1 / (u^2 + 5)^2 ] du
This integral can be evaluated using a basic integration formula or partial fraction decomposition, and the final answer can be obtained by substituting back u = tan(θ).
(ii) To evaluate the integral [ x^3 / (x^2 + 4)^2 ] using regular substitution, we can make the substitution u = x^2 + 4.
First, let's find the derivative of u with respect to x:
du/dx = 2x
Next, we can replace x^3 and dx in the integral with expressions involving u:
x^3 = (u - 4)^3/2
dx = (1/2x) du
Now let's substitute these expressions into the original integral:
∫ [ x^3 / (x^2 + 4)^2 ] dx = ∫ [ (u - 4)^3/2 / u^2 ] (1/2x) du
= (1/2)∫ (u - 4)^3/2 / u^2 du
At this point, we can simplify the integrand and then evaluate the integral using basic integration rules.
(iii) To check that the answers obtained in parts (i) and (ii) are the same, we can use a right triangle.
Consider a right triangle with one acute angle θ and sides of length x and (x^2 + 4). By the definition of trigonometric ratios, we have:
sin(θ) = x / (x^2 + 4) and cos(θ) = sqrt(u) / (x^2 + 4).
From the triangle, we can also find the value of tan(θ):
tan(θ) = sin(θ) / cos(θ) = (x / (x^2 + 4)) / (sqrt(u) / (x^2 + 4)) = x / sqrt(u).
Substituting this result back into the trigonometric substitution (i), we get:
x = 2tan(θ) = 2(x / sqrt(u)) = 2x / sqrt(u),
Simplifying, we find: x^2 = 4 / u.
Now let's compare the integrands obtained in parts (i) and (ii) using these trigonometric relationships.
In part (i), after applying the trigonometric substitution, we obtained the integral 2∫ [ secθtan(θ) / (1 + tan^2(θ) + 4)^2 ] dθ.
Using the relationships x^2 = 4 / u and x = 2x / sqrt(u), we can rewrite this integral in terms of x and u.
After making these substitutions, we can evaluate the integral and obtain the final answer.
Similarly, in part (ii), after applying the regular substitution, we obtained the integral (1/2)∫ (u - 4)^3/2 / u^2 du.
We can substitute x^2 = 4 / u into this integral as well, and evaluate to obtain the final answer.
By obtaining the same final answer in both cases, we can conclude that the trigonometric substitution and regular substitution methods yield equivalent results for this integral.