Answer:
To evaluate the given integrals using the indicated trigonometric substitutions, we'll first substitute the given expressions, sketch the associated right triangles, and then integrate with respect to the new variable. Here are the solutions for each of the integrals:
5. ∫(1 - x^2)x^3 dx, with x = sin(θ):
We need to find dx in terms of dθ:
dx = cos(θ) dθ
Now, substitute:
∫(1 - sin^2(θ))(sin^3(θ) cos(θ)) dθ
Simplify:
∫(cos(θ) - sin^2(θ) cos(θ)) sin^3(θ) dθ
∫(cos(θ)sin^3(θ) - sin^5(θ)cos(θ)) dθ
Now, integrate term by term:
∫cos(θ)sin^3(θ) dθ - ∫sin^5(θ)cos(θ) dθ
The first integral can be solved using a simple u-substitution, where u = sin(θ):
Let u = sin(θ), du = cos(θ) dθ
So the integral becomes:
∫u^3 du
Integrate:
(1/4)u^4 + C1
Now, for the second integral, we can use a similar u-substitution:
Let u = sin(θ), du = cos(θ) dθ
So the integral becomes:
-∫u^5 du
Integrate:
-(1/6)u^6 + C2
Finally, substitute back for u:
(1/4)sin^4(θ) - (1/6)sin^6(θ) + C
6. ∫(9 + x^2)x^3 dx, with x = 3tan(θ):
We need to find dx in terms of dθ:
dx = 3sec^2(θ) dθ
Now, substitute:
∫(9 + (3tan(θ))^2)(3tan(θ))^3 (3sec^2(θ)) dθ
Simplify:
∫(9 + 9tan^2(θ))(27tan^3(θ)sec^2(θ)) dθ
∫(9*27tan^3(θ)sec^2(θ) + 9*27tan^5(θ)sec^2(θ)) dθ
Now, integrate term by term:
9*27∫tan^3(θ)sec^2(θ) dθ + 9*27∫tan^5(θ)sec^2(θ) dθ
The first integral can be solved using a simple u-substitution, where u = tan(θ):
Let u = tan(θ), du = sec^2(θ) dθ
So the integral becomes:
9*27∫u^3 du
Integrate:
9*27(1/4)u^4 + C1
Now, for the second integral, we can use a similar u-substitution:
Let u = tan(θ), du = sec^2(θ) dθ
So the integral becomes:
9*27∫u^5 du
Integrate:
9*27(1/6)u^6 + C2
Finally, substitute back for u:
9*27(1/4)tan^4(θ) + 9*27(1/6)tan^6(θ) + C
7. ∫x^4 / (x^2 - 25) dx, with x = 25sec(θ):
We need to find dx in terms of dθ:
dx = 25sec(θ)tan(θ) dθ
Now, substitute:
∫(25sec(θ))^4 / ((25sec(θ))^2 - 25) (25sec(θ)tan(θ)) dθ
Simplify:
∫(25^4sec^4(θ)) / (25^2sec^2(θ) - 25) (25sec(θ)tan(θ)) dθ
∫(25^4sec^4(θ)) / (25^2(sec^2(θ) - 1)) (25sec(θ)tan(θ)) dθ
∫(25^4sec^4(θ)) / (25^2tan^2(θ)) (25sec(θ)tan(θ)) dθ
Now, simplify further:
∫25^2sec^3(θ) dθ
This integral can be evaluated using a basic trigonometric identity:
∫25^2sec^3(θ) dθ = (25^2/2)sec(θ)tan(θ) + C
Substitute back for x:
(25^2/2)sec(θ)tan(θ) + C
8. ∫x^2 / (2 - x^2) dx, with x = 2sin(θ):
We need to find dx in terms of dθ:
dx = 2cos(θ) dθ
Now, substitute:
∫(2sin(θ))^2 / (2 - (2sin(θ))^2) (2cos(θ)) dθ
Simplify:
∫(4sin^2(θ)) / (2 - 4sin^2(θ)) (2cos(θ)) dθ
∫(4sin^2(θ)) / (2cos^2(θ)) (2cos(θ)) dθ
Now, simplify further:
∫4sin^2(θ) dθ
This integral can be evaluated using a trigonometric identity:
4sin^2(θ) = 2 - 2cos(2θ)
So, the integral becomes:
∫(2 - 2cos(2θ)) dθ
Integrate term by term:
2θ - sin(2θ) + C
Substitute back for x:
2sin^(-1)(x/2) - sin(2sin^(-1)(x/2)) + C
These are the results for the given integrals using the indicated trigonometric substitutions.
Explanation: