Integration by Parts in Theory: Identities, Proofs, and Recurrences

Integration by parts is derived from the product rule for differentiation. It transforms the integral of a product of functions into a simpler integral, allowing us to find the antiderivative.

Using the integration by parts formula for definite integrals: ∫₁⁴ f(x)g'(x) dx = [f(x)g(x)]₁⁴ - ∫₁⁴ g(x)f'(x) dx. First, evaluate the boundary term: f(4)g(4) - f(1)g(1) = (6)(3) - (2)(1) = 18 - 2 = 16. We are given that ∫₁⁴ g(x)f'(x) dx = 10. Therefore, the desired integral is 16 - 10 = 6.

The student made a sign error in the final step. Using parts with u = x and dv = cos(x) dx, we get v = sin(x). The formula gives x sin(x) - ∫ sin(x) dx. The integral of sin(x) is -cos(x). Therefore, the result should be x sin(x) - (-cos(x)) + C, which simplifies to x sin(x) + cos(x) + C. The student failed to distribute the negative sign.

In the integration by parts of x e^{kx}, the term v is found by integrating e^{kx}, which results in (1/k)e^{kx}. This coefficient appears in the first term of the solution: uv = x(1/k)e^{kx}. Comparing x(1/k)e^{kx} with the given solution (x/3)e^{3x}, we can see that 1/k = 1/3, which implies k = 3.

Using integration by parts, let u = arcsin(x) and dv = dx. Then du = 1/√(1-x²) dx and v = x. Thus, we obtain ∫ arcsin(x) dx = x arcsin(x) - ∫ x/√(1-x²) dx. To solve the remaining integral, let w = 1-x², so dw = -2x dx, or x dx = -dw/2. We have ∫ x/√(1-x²) dx = ∫ (-½)w^{-1/2} dw = -w^{1/2} = -√(1-x²). Substituting back, we obtain x arcsin(x) - (-√(1-x²)) + C = x arcsin(x) + √(1-x²) + C.

Use integration by parts with u = x, dv = sin x dx, du = dx, v = -cos x. Then ∫ x sin x dx = -x cos x + ∫ cos x dx = -x cos x + sin x + C. Plug in the limits: at π/2, - (π/2)·cos(π/2) + sin(π/2) = 0 + 1 = 1. At 0, -0·cos 0 + sin 0 = 0. The difference is 1.

The integration by parts formula states ∫ u dv = uv - ∫ v du. Rearranging this, we get ∫ v du = uv - ∫ u dv. We are given that uv = x² and ∫ u dv = x² - eˣ. Substituting these into the rearranged formula: ∫ v(x) u'(x) dx = x² - (x² - eˣ) = x² - x² + eˣ = eˣ + C.

In integration by parts, we must be able to find v by integrating dv. If we choose dv = ln(x) dx, finding v requires knowing the integral of ln(x) (which itself requires integration by parts). The standard strategy (LIATE) suggests choosing u = ln(x) because it simplifies to 1/x when differentiated, while dv = x dx is easy to integrate.

Set u = (ln x)², dv = dx, giving du = (2 ln x)/x dx, v = x. Integration by parts yields x (ln x)² - ∫ 2 ln x dx. Evaluate the remaining integral by parts again: let u = ln x, dv = dx, du = (1/x) dx, v = x. Then ∫ ln x dx = x ln x - ∫ dx = x ln x - x + C. Substituting back gives x (ln x)² - 2(x ln x - x) + C = x (ln x)² - 2x ln x + 2x + C.

Use the antiderivative from the previous problem: ∫ x² eˣ dx = eˣ (x² - 2x + 2) + C. Evaluate at the upper limit 1: e (1 - 2 + 2) = e. Evaluate at the lower limit 0: e⁰(0 - 0 + 2) = 2. Subtract to obtain e - 2.

When applying integration by parts the first time, we get uv - ∫ v du, which gives a positive x²eˣ term and a negative integral. When applying it a second time to the remaining integral, the formula again generates a subtraction (A - (B - C)). Distributing the negative sign from the first step results in A - B + C. Therefore, the terms x²eˣ, 2xeˣ, and 2eˣ have the signs positive, negative, and positive, respectively.

Use integration by parts with u = (ln x)ⁿ, dv = dx, du = n (ln x)ⁿ⁻¹ (1/x) dx, v = x. Then Iₙ = x (ln x)ⁿ - ∫ x * n (ln x)ⁿ⁻¹ (1/x) dx = x (ln x)ⁿ - n ∫ (ln x)ⁿ⁻¹ dx = x (ln x)ⁿ - n Iₙ₋₁.

Set u = arctan x, dv = dx, du = 1/(1 + x²) dx, v = x. Integration by parts yields x arctan x - ∫ x/(1 + x²) dx. The remaining integral is solved by substitution w = 1 + x², dw = 2x dx, giving (½) ∫ dw/w = (½) ln|w| + C = (½) ln(1 + x²) + C. The result is x arctan x - (½) ln(1 + x²) + C.

Let u = arccos(2x). Using the chain rule, du = -1/√(1-(2x)²) * 2 dx = -2/√(1-4x²) dx. Let dv = dx, so v = x. The formula is uv - ∫ v du. The integral part (∫ v du) is ∫ x * (-2/√(1-4x²)) dx = ∫ -2x / √(1-4x²) dx.

To find the recurrence relation for Iₙ+1, we evaluate the integral of xⁿ⁺¹ eˣ dx. We use integration by parts with u = xⁿ⁺¹ and dv = eˣ dx. Differentiating u gives du = (n+1)xⁿ dx, and integrating dv gives v = eˣ. Applying the integration by parts formula ∫ u dv = uv - ∫ v du, we get xⁿ⁺¹ eˣ - ∫ (n+1)xⁿ eˣ dx. We can factor out the constant (n+1) from the integral, leaving xⁿ⁺¹ eˣ - (n+1)∫ xⁿ eˣ dx. Since ∫ xⁿ eˣ dx is defined as Iₙ, the equation becomes Iₙ+1 = xⁿ⁺¹ eˣ - (n+1)Iₙ.