Integration by Parts with Exponentials and Trigonometric Factors

Let u = x + 1 and dv = eˣ dx. Then du = dx and v = eˣ. Applying integration by parts: ∫ (x + 1) eˣ dx = (x + 1)eˣ - ∫ eˣ dx = (x + 1)eˣ - eˣ + C = eˣ (x + 1 - 1) + C = x eˣ + C.

Let u = x and dv = cos(2x) dx. Then du = dx and v = (½) sin(2x). Applying integration by parts: ∫ x cos(2x) dx = x × (½) sin(2x) - ∫ (½) sin(2x) dx = (x/2) sin(2x) - (½) ∫ sin(2x) dx. The integral of sin(2x) is -(½) cos(2x). So we get (x/2) sin(2x) - (½) × [-(½) cos(2x)] + C = (x/2) sin(2x) + (1/4) cos(2x) + C.

Integration by parts comes from the product rule for differentiation. The product rule states that d(uv)/dx = u dv/dx + v du/dx, which can be written as d(uv) = u dv + v du. Integrating both sides gives uv = ∫ u dv + ∫ v du. Rearranging this equation yields the integration by parts formula: ∫ u dv = uv - ∫ v du. This shows that integration by parts is essentially the inverse operation of the product rule.

First application: Let u = x² and dv = sin x dx, so du = 2x dx and v = -cos x. Then ∫ x² sin x dx = -x² cos x - ∫ (-cos x)(2x dx) = -x² cos x + 2 ∫ x cos x dx. Second application on ∫ x cos x dx: Let u = x, dv = cos x dx, so du = dx, v = sin x. This gives ∫ x cos x dx = x sin x - ∫ sin x dx = x sin x + cos x + C. Substituting back: ∫ x² sin x dx = -x² cos x + 2(x sin x + cos x) + C = -x² cos x + 2x sin x + 2 cos x + C.

First find the antiderivative. Let u = x, dv = cos x dx, so du = dx, v = sin x. Then ∫ x cos x dx = x sin x - ∫ sin x dx = x sin x + cos x + C. Now evaluate from 0 to π/2: [(π/2) sin(π/2) + cos(π/2)] - [0 sin(0) + cos(0)] = [(π/2) × 1 + 0] - [0 + 1] = π/2 - 1.

First, simplify ln(x²) = 2 ln x using logarithm properties. So ∫ ln(x²) dx = ∫ 2 ln x dx = 2 ∫ ln x dx. Now evaluate ∫ ln x dx using integration by parts: Let u = ln x, dv = dx, so du = (1/x) dx, v = x. Then ∫ ln x dx = x ln x - ∫ x × (1/x) dx = x ln x - ∫ dx = x ln x - x + C. Therefore, 2 ∫ ln x dx = 2(x ln x - x) + C = 2x ln x - 2x + C. Since ln(x²) = 2 ln x, we can also write this as x ln(x²) - 2x + C.

The correct choice is u = x and dv = sin x dx. This choice works well because the derivative of x is 1 (simplifying the polynomial part), and the antiderivative of sin x is -cos x (easy to find). If we chose u = sin x and dv = x dx, we would get du = cos x dx and v = x²/2, leading to a more complicated integral ∫ (x²/2) cos x dx that is harder than the original.

First application: Let u = cos x, dv = eˣ dx, so du = -sin x dx, v = eˣ. Then ∫ eˣ cos x dx = eˣ cos x - ∫ eˣ (-sin x) dx = eˣ cos x + ∫ eˣ sin x dx. Second application on ∫ eˣ sin x dx: Let u = sin x, dv = eˣ dx, so du = cos x dx, v = eˣ. This gives ∫ eˣ sin x dx = eˣ sin x - ∫ eˣ cos x dx. Substituting back: ∫ eˣ cos x dx = eˣ cos x + [eˣ sin x - ∫ eˣ cos x dx] = eˣ cos x + eˣ sin x - ∫ eˣ cos x dx. Now add ∫ eˣ cos x dx to both sides: 2∫ eˣ cos x dx = eˣ(cos x + sin x). Therefore, ∫ eˣ cos x dx = (eˣ)(sin x + cos x)/2 + C.

Let u = x and dv = e⁻ˣ dx. Then du = dx and v = -e⁻ˣ (since the integral of e⁻ˣ is -e⁻ˣ). Applying integration by parts: ∫ x e⁻ˣ dx = x × (-e⁻ˣ) - ∫ (-e⁻ˣ) dx = -x e⁻ˣ + ∫ e⁻ˣ dx. The integral of e⁻ˣ is -e⁻ˣ. So we get -x e⁻ˣ + (-e⁻ˣ) + C = -x e⁻ˣ - e⁻ˣ + C. This can also be factored as -e⁻ˣ(x + 1) + C.

For integrals of the form ∫ xⁿ eˣ dx, you typically need to apply integration by parts n times. Each application reduces the power of x by 1. For example, with x² eˣ, you apply it twice: first to reduce x² to x, then again to reduce x to a constant. After n applications, you'll be left with ∫ eˣ dx, which is straightforward to evaluate.

Let u = ln x and dv = x² dx, so du = (1/x) dx and v = x³/3. Then ∫ x² ln x dx = (x³/3) ln x - ∫ (x³/3)(1/x) dx = (x³/3) ln x - ∫ (x²/3) dx = (x³/3) ln x - x³/9 + C. Now evaluate from 1 to 2: [(8/3) ln 2 - 8/9] - [(⅓) ln 1 - 1/9] = [(8/3) ln 2 - 8/9] - [0 - 1/9] = (8/3) ln 2 - 8/9 + 1/9 = (8/3) ln 2 - 7/9.

Integration by parts is based on the product rule for differentiation (d(uv) = u dv + v du), while u-substitution is based on the chain rule for differentiation. Integration by parts is used when the integrand is a product of two functions where one function is easily differentiable and the other is easily integrable. U-substitution is used when the integrand contains a function and its derivative.

First application: Let u = x², dv = e²ˣ dx, so du = 2x dx, v = (½)e²ˣ. Then ∫ x² e²ˣ dx = x² × (½)e²ˣ - ∫ (½)e²ˣ × 2x dx = (x²/2)e²ˣ - ∫ x e²ˣ dx. Second application on ∫ x e²ˣ dx: Let u = x, dv = e²ˣ dx, so du = dx, v = (½)e²ˣ. This gives ∫ x e²ˣ dx = x × (½)e²ˣ - ∫ (½)e²ˣ dx = (x/2)e²ˣ - (½) × (½)e²ˣ = (x/2)e²ˣ - (1/4)e²ˣ. Substituting back: ∫ x² e²ˣ dx = (x²/2)e²ˣ - [(x/2)e²ˣ - (1/4)e²ˣ] + C = (x²/2)e²ˣ - (x/2)e²ˣ + (1/4)e²ˣ + C.

We add the constant of integration (+ C) because integration is the reverse process of differentiation. When we differentiate any constant, we get zero. Therefore, when we integrate a function, we get a family of functions that differ by a constant. The + C represents this family of antiderivatives. It's essential for indefinite integrals but not needed for definite integrals since the constant cancels out when evaluating at the limits.

This requires integration by parts three times. First: u = x³, dv = eˣ dx → du = 3x² dx, v = eˣ → ∫ x³ eˣ dx = x³ eˣ - 3 ∫ x² eˣ dx. Second: For ∫ x² eˣ dx, u = x², dv = eˣ dx → du = 2x dx, v = eˣ → ∫ x² eˣ dx = x² eˣ - 2 ∫ x eˣ dx. Third: For ∫ x eˣ dx, u = x, dv = eˣ dx → du = dx, v = eˣ → ∫ x eˣ dx = x eˣ - ∫ eˣ dx = x eˣ - eˣ. Working backwards: ∫ x² eˣ dx = x² eˣ - 2(x eˣ - eˣ) = x² eˣ - 2x eˣ + 2eˣ. Then ∫ x³ eˣ dx = x³ eˣ - 3(x² eˣ - 2x eˣ + 2eˣ) = x³ eˣ - 3x² eˣ + 6x eˣ - 6eˣ + C = eˣ (x³ - 3x² + 6x - 6) + C.