9–61. Trigonometric integrals Evaluate the following integrals.
57. ∫ from 0 to π of (1 - cos2x)³ᐟ² dx

Verified step by step guidance

1. Step 1: Recognize that the integral involves a trigonometric expression raised to a fractional power. To simplify, use the trigonometric identity: $cos(2x)\; =\; 1\; -\; 2sin²(x)$. Rewrite $1\; -\; cos(2x)$ as $2sin²(x)$.
2. Step 2: Substitute $1\; -\; cos(2x)$ with $2sin²(x)$ in the integral. The integral becomes: $\int \; from\; 0\; to\; \pi \; of\; (2sin²(x))³ᐟ²\; dx$.
3. Step 3: Simplify the expression $(2sin²(x))³ᐟ²$. This can be rewritten as $(2³ᐟ²)(sin²(x))³ᐟ²$, where $2³ᐟ²$ is a constant factor that can be factored out of the integral.
4. Step 4: Rewrite the integral as $2³ᐟ²\; \int \; from\; 0\; to\; \pi \; of\; (sin(x))²\; dx$. To evaluate this integral, use the power-reduction formula for sine: $sin²(x)\; =\; (1\; -\; cos(2x))/2$. Substitute this into the integral.
5. Step 5: After substitution, the integral becomes $2³ᐟ²\; \int \; from\; 0\; to\; \pi \; of\; (1\; -\; cos(2x))/2\; dx$. Split the integral into two parts: $(2³ᐟ²/2)\; \int \; from\; 0\; to\; \pi \; of\; 1\; dx$ and $-(2³ᐟ²/2)\; \int \; from\; 0\; to\; \pi \; of\; cos(2x)\; dx$. Evaluate each part separately using basic integration rules.

Key Concepts

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are fundamental in calculus, particularly in integrals involving periodic functions. Understanding their properties, such as periodicity and symmetry, is crucial for evaluating integrals that include these functions. In this case, the integral involves the cosine function, which can be manipulated using trigonometric identities.

Integration Techniques

Integration techniques, including substitution and integration by parts, are essential for solving complex integrals. In this problem, recognizing the structure of the integrand (1 - cos(2x))^3/2 may suggest a substitution or a trigonometric identity to simplify the integral. Mastery of these techniques allows for the effective evaluation of integrals that may initially appear daunting.

Definite Integrals

Definite integrals calculate the area under a curve between specified limits, in this case, from 0 to π. Understanding the properties of definite integrals, such as the Fundamental Theorem of Calculus, is vital for evaluating the integral accurately. This theorem connects differentiation and integration, allowing for the evaluation of the integral by finding an antiderivative and applying the limits.