Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.
∫₀^π/2 x³ cos 2x dx

Verified step by step guidance

1. Identify the integral to solve: \(\int_0^{\frac{\pi}{2}} x^3 \cos(2x) \, dx\).
2. Recognize that this integral involves a product of a polynomial (\(x^3\)) and a trigonometric function (\(\cos(2x)\)), which suggests using integration by parts.
3. Set up integration by parts by choosing \(u = x^3\) (which simplifies when differentiated) and \(dv = \cos(2x) \, dx\) (which can be integrated easily). Recall the formula: \(\int u \, dv = uv - \int v \, du\).
4. Compute \(du = 3x^2 \, dx\) and find \(v\) by integrating \(dv\): \(v = \int \cos(2x) \, dx = \frac{1}{2} \sin(2x)\).
5. Apply the integration by parts formula: \(\int_0^{\frac{\pi}{2}} x^3 \cos(2x) \, dx = \left. x^3 \cdot \frac{1}{2} \sin(2x) \right|_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} \frac{1}{2} \sin(2x) \cdot 3x^2 \, dx\). This reduces the original integral to a new integral involving \(x^2 \sin(2x)\), which may require repeating integration by parts.

Key Concepts

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It is based on the product rule for differentiation and follows the formula ∫u dv = uv - ∫v du. Choosing u and dv wisely simplifies the integral, especially when one function becomes simpler upon differentiation.

Definite Integrals

Definite integrals calculate the net area under a curve between two limits. They produce a numerical value and require evaluating the antiderivative at the upper and lower bounds. Understanding how to apply limits after integration is essential for solving definite integrals.

Trigonometric Functions in Integration

Integrating functions involving trigonometric terms often requires using identities or special techniques. Recognizing how to handle integrals with cosine or sine, especially when multiplied by polynomials, helps in simplifying and solving the integral effectively.