We recognize this as a gamma function integral. The gamma function is defined as:
\[ \Gamma(n) = \int_{0}^{\infty} x^{n-1} e^{-x} \, dx \]
The gamma function generalizes the factorial function to non-integer values.
First, let's make a substitution to match the gamma function form. Let u = 3x, then du = 3dx and x = u/3.
This substitution helps us match the standard gamma function form.
Substitute into the integral:
\[ \int_{0}^{\infty} \left(\frac{u}{3}\right)^5 e^{-u} \frac{du}{3} = \frac{1}{3^6} \int_{0}^{\infty} u^5 e^{-u} \, du \]
Notice how the substitution transforms the integral into a standard gamma function multiplied by a constant.
Now we recognize the gamma function Γ(6) since Γ(n) = (n-1)!:
\[ \frac{1}{3^6} \Gamma(6) = \frac{5!}{3^6} \]
For positive integers n, Γ(n) = (n-1)!, which connects to factorial concepts you already know.
Calculate the final value:
\[ \frac{120}{729} = \frac{40}{243} \]
Let's make a substitution: u = αx², then du = 2αx dx and x² = u/α.
This substitution simplifies the exponential term to e⁻ᵘ.
Rewrite the integral:
\[ \int_{0}^{\infty} e^{-u} x^3 \, dx = \int_{0}^{\infty} e^{-u} (x^2) (x dx) = \int_{0}^{\infty} e^{-u} \left(\frac{u}{\alpha}\right) \left(\frac{du}{2\alpha}\right) \]
We're breaking x³ into x²·x to match our substitution terms.
Simplify:
\[ \frac{1}{2\alpha^2} \int_{0}^{\infty} u e^{-u} \, du \]
The integral now clearly shows a gamma function Γ(2).
We recognize this as Γ(2) = 1! = 1:
\[ \frac{1}{2\alpha^2} = 32 \]
Now we can solve for α using basic algebra.
Solve for α:
\[ \frac{1}{2\alpha^2} = 32 \implies \alpha^2 = \frac{1}{64} \implies \alpha = \frac{1}{8} \]
Let's make the substitution u = tan(x), then du = sec²(x) dx = dx/cos²(x).
The tangent substitution is natural here because of the e⁻ᵗᵃⁿˣ term.
Rewrite the integral in terms of u:
\[ \int_{0}^{\infty} e^{-u} \cdot \frac{1}{\cos^4 x} \, du \]
We still need to express cos⁴x in terms of u.
Express cos²x in terms of u: cos²x = 1/(1 + tan²x) = 1/(1 + u²)
So 1/cos⁴x = (1 + u²)²
This trigonometric identity is crucial for the simplification.
Now the integral becomes:
\[ \int_{0}^{\infty} e^{-u} (1 + u^2)^2 \, du \]
The integral now looks much simpler in terms of u.
Expand the integrand:
\[ \int_{0}^{\infty} e^{-u} (1 + 2u^2 + u^4) \, du \]
We can now split this into three separate integrals.
This can be separated into gamma functions:
\[ Γ(1) + 2Γ(3) + Γ(5) = 0! + 2×2! + 4! = 1 + 4 + 24 = 29 \]