1. integral (x^3 * sqrt(x^2-4)) dx
I'm confused on how to start this problem, if I am doing trig substitution.
No need for trig substitution. We'll just do normal substitution:
Let u = sqrt(x^2 - 4)
Thus, we can make the following relation:
u^2 = x^2 - 4
x^2 = u^2 + 4
2u du = 2x dx
dx = u/x du
Substituting,
∫ ( x^3 * sqrt(x^2-4) ) dx
∫ ( x^3 * u * u / x ) du
∫ ( x^2 * u^2 ) du
∫ ( (u^2 + 4) u^2 ) du
∫ ( u^4 + 4u^2 ) du
= (1/5)u^5 + (4/3)u^3 + C
Substituting back the expression for u:
= (1/5) [sqrt(x^2 - 4)]^5 + (4/3) [sqrt(x^2-4)]^3 + C
= (1/5)(x^2 - 4)^5/2 + (4/3)(x^2-4)^3/2 + C
You can further simplify this if we want, by first factoring the common factor, (x^2-4)^(3/2):
= (x^2-4)^(3/2) * [ (1/5)*(x^2-4) + 4/3 ] + C
Now, do the rest if you want to continue this simplification.
hope this helps~ `u`
To solve the integral ∫(x^3 * √(x^2-4)) dx using trigonometric substitution, you'll need to follow these steps:
1. Identify the appropriate trigonometric substitution: In this case, since the quadratic term involved is x^2-4, which resembles the identity of a^2 - x^2 (where a is a constant), we can use the substitution x = 2secθ.
2. Determine the appropriate differentials: In this case, dx can be replaced by 2secθ * tanθ dθ, thanks to the derivative of secant.
3. Substitute the trigonometric expression: Replace x with 2secθ, and dx with 2secθ * tanθ dθ in the integral.
Therefore, the integral becomes:
∫(2secθ)^3 * √((2secθ)^2-4) * 2secθ * tanθ dθ.
4. Simplify the expression: Simplify further by expanding the cube and simplifying the square root term.
5. Combine like terms: Combine any similar terms, such as multiplying/sec^3θ * secθ * tanθ.
6. Evaluate the integral: Apply the appropriate trigonometric identities to simplify the expression further, if necessary. Finally, integrate the simplified expression.
Remember, these are general steps for trigonometric substitution. Always check the domain and ensure the substitution is appropriate for the given integral.