Advanced Integration Techniques

Example:

Integrate ∫ x e^x dx.

Let u = x and dv = e^x dx. Then, du = dx and v = e^x.

Using the integration by parts formula:

∫ x e^x dx = x e^x - ∫ e^x dx

= x e^x - e^x + C

Therefore, the integral is x e^x - e^x + C.

Example:

Integrate ∫ sin²(x) dx.

Use the identity sin²(x) = (1 - cos(2x))/2.

Then, ∫ sin²(x) dx = ∫ (1 - cos(2x))/2 dx

= 1/2 ∫ (1 - cos(2x)) dx

= 1/2 (x - 1/2 sin(2x)) + C

= x/2 - 1/4 sin(2x) + C

Therefore, the integral is x/2 - 1/4 sin(2x) + C.

Example:

Integrate ∫ dx / √(a² - x²).

Use the substitution x = a sin(θ), dx = a cos(θ) dθ, and √(a² - x²) = a cos(θ).

Then, ∫ dx / √(a² - x²) = ∫ a cos(θ) dθ / a cos(θ)

= ∫ dθ

= θ + C

Since x = a sin(θ), θ = arcsin(x/a).

Therefore, the integral is arcsin(x/a) + C.

Example:

Integrate ∫ (2x + 3) / (x² - x - 2) dx.

Factor the denominator: x² - x - 2 = (x - 2)(x + 1).

Express the integrand as a sum of partial fractions:

(2x + 3) / ((x - 2)(x + 1)) = A / (x - 2) + B / (x + 1).

Multiply through by the denominator to find A and B:

2x + 3 = A(x + 1) + B(x - 2).

Set up a system of equations by equating coefficients:

A + B = 2

A - 2B = 3

Solve the system to find A = 1 and B = 1.

Therefore, the integral is ∫ 1 / (x - 2) dx + ∫ 1 / (x + 1) dx

= ln|x - 2| + ln|x + 1| + C

Example:

Evaluate the improper integral ∫₁^∞ 1/x² dx.

Rewrite the integral with a limit:

∫₁^∞ 1/x² dx = lim_t→∞ ∫₁^t 1/x² dx

Evaluate the integral:

∫ 1/x² dx = -1/x

Then, lim_t→∞ [-1/x]₁^t

= lim_t→∞ (-1/t + 1)

= 1

Therefore, the value of the improper integral is 1.