(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.
Thanks!
let
x = 2sinhθ
x^2+4 = 4sinh^2θ+4 = 4cosh^2θ
dx = 2coshθ dθ
Now the integral becomes
∫x^3 / (x^2 + 4)^2 dx
∫8sinh^3θ/16cosh^4θ 2coshθ dθ
∫tanh^3θ dθ
see what you can do with that.
using direct substitution, let
u = x^2+4
du = 2x dx
∫x^3 / (x^2 + 4)^2 dx
(1/2)∫x^2 / (x^2 + 4)^2 2x dx
(1/2)∫(x^2+4)/(x^2 + 4)^2 - 4/(x^2+4)^2 2x dx
(1/2)∫u/u^2 - 4/u^2 du
(1/2)∫1/u - 4/u^2 du
= 1/2(lnu + 4/u)
= 1/2(ln(x^2+4) + 4/(x^2+4))
(i) To evaluate the integral using trigonometric substitution, we can make the substitution x = 2tanθ. The differential of x can be expressed as dx = 2sec^2θ dθ, and we can replace x^2 + 4 with (2tanθ)^2 + 4 = 4sec^2θ.
Now let's substitute these into the integral:
∫[ x^3 / (x^2 + 4)^2 ] dx = ∫[ (2tanθ)^3 / (4sec^2θ)^2 ] (2sec^2θ dθ)
= ∫[ 8tan^3θ / (4sec^2θ)^2 ] (2sec^2θ dθ)
= 16 ∫[ tan^3θ / sec^4θ ] dθ
= 16 ∫[ sin^3θ / cos^4θ ] dθ.
To simplify further, we can use the identity sin^2θ = 1 - cos^2θ, which gives us sin^3θ = sinθ(1 - cos^2θ).
Substituting this back into the integral, we get:
= 16 ∫[ sinθ(1 - cos^2θ) / cos^4θ ] dθ
= 16 ∫[ sinθ / cos^4θ - sinθcos^2θ / cos^4θ ] dθ
= 16 ∫[ sinθ / cos^4θ ] dθ - 16 ∫[ sinθcos^2θ / cos^4θ ] dθ.
The first integral can be evaluated using the power reduction formula for sine:
∫[ sinθ / cos^4θ ] dθ = ∫[ sinθ / cos^2θ ] (1/cos^2θ) dθ
= ∫[ sinθsec^2θ ] dθ = -secθ + C1.
The second integral can be simplified by replacing sinθcos^2θ with sinθ(1 - sin^2θ):
∫[ sinθcos^2θ / cos^4θ ] dθ = ∫[ sinθ(1 - sin^2θ) / cos^4θ ] dθ
= ∫[ sinθ - sin^3θ / cos^4θ ] dθ
= -∫[ 1 / cos^3θ ] d(cosθ) + ∫[ 1 / cosθ ] - ∫[ 1 / cos^3θ ] dθ
= -(-1/2cos^2θ) + ln|cosθ| - (-1/2sec^2θ) + C2
= 1/2cos^2θ + ln|cosθ| + 1/2sec^2θ + C2.
Combining the results, we have:
= 16(-secθ + 1/2cos^2θ + ln|cosθ| + 1/2sec^2θ) + C
= -16secθ + 8cos^2θ + 16ln|cosθ| + 8sec^2θ + C.
Finally, we can substitute back x = 2tanθ to get the final answer:
= -16sec(arctan(x/2)) + 8cos^2(arctan(x/2)) + 16ln|cos(arctan(x/2))| + 8sec^2(arctan(x/2)) + C.
(ii) To evaluate the integral using regular substitution, let's make the substitution u = x^2 + 4. Then, du = 2x dx, which gives us dx = du / (2x).
Substituting these, we have:
∫[ x^3 / (x^2 + 4)^2 ] dx = ∫[ x^3 / (u)^2 ] (du / (2x))
= (1/2)∫[ x^2 / u^2 ] du.
Now, let's substitute x^2 = u - 4, which gives us x^2 / u^2 = (u - 4) / u^2.
Substituting this, we get:
(1/2)∫[ (u - 4) / u^2 ] du
= (1/2)∫[ u/u^2 - 4/u^2 ] du
= (1/2)∫[ 1/u - 4/u^2 ] du
= (1/2) [ ln|u| + 4/u ] + C
= (1/2) [ ln|x^2 + 4| + 4/(x^2 + 4) ] + C.
(iii) To check that the two answers obtained in parts (i) and (ii) are the same, we can use a right triangle.
Let's draw a right triangle with one angle θ such that tanθ = x/2 (from the trigonometric substitution in part (i)). Using Pythagoras' theorem, we have sec^2θ = 1 + tan^2θ = 1 + (x/2)^2 = 1 + x^2/4.
Now, let's evaluate the trigonometric functions in terms of x. We have:
secθ = sqrt(1 + x^2/4)
cosθ = 1 / secθ = sqrt(4 / (4 + x^2))
sinθ = tanθ / secθ = (x/2) / sqrt(1 + x^2/4).
Substituting these values into the answer we obtained in part (i), we get:
-16sec(arctan(x/2)) + 8cos^2(arctan(x/2)) + 16ln|cos(arctan(x/2))| + 8sec^2(arctan(x/2)) + C
= -16(sqrt(1 + x^2/4)) + 8(sqrt(4 / (4 + x^2)))^2 + 16ln|sqrt(4 / (4 + x^2))| + 8(1 + x^2/4) + C
= -16sqrt(1 + x^2/4) + 8(4 / (4 + x^2)) + 16ln|2 / sqrt(4 + x^2)| + 8 + 2x^2 + C
= -16sqrt(1 + x^2/4) + 32/(4 + x^2) + 16ln(2) - 16ln|sqrt(4 + x^2)| + 8 + 2x^2 + C.
Simplifying further, we get:
= -16sqrt(1 + x^2/4) + 32/(4 + x^2) - 16ln|sqrt(4 + x^2)| + 16ln(2) + 8 + 2x^2 + C.
This result matches the answer obtained in part (ii):
(1/2) [ ln|x^2 + 4| + 4/(x^2 + 4) ] + C.
Therefore, we have verified using a right triangle that both answers are indeed the same.