7–84. Evaluate the following integrals.
7. ∫ from 0 to π/2 [sin θ / (1 + cos² θ)] dθ

Verified step by step guidance

1. Step 1: Recognize that the integral involves trigonometric functions. The integrand is sin(θ) / (1 + cos²(θ)). To simplify, consider substitution methods or trigonometric identities.
2. Step 2: Use the substitution u = cos(θ). Then, compute du = -sin(θ)dθ. This substitution will transform the integral into terms of u, making it easier to evaluate.
3. Step 3: Change the limits of integration according to the substitution. When θ = 0, u = cos(0) = 1. When θ = π/2, u = cos(π/2) = 0. The integral now has limits from u = 1 to u = 0.
4. Step 4: Rewrite the integral in terms of u. The original integral becomes ∫ from 1 to 0 [-du / (1 + u²)]. Notice the negative sign from du = -sin(θ)dθ.
5. Step 5: Reverse the limits of integration to remove the negative sign, resulting in ∫ from 0 to 1 [du / (1 + u²)]. This integral is now in a standard form, which can be evaluated using the arctangent function: ∫ [du / (1 + u²)] = arctan(u).

Key Concepts

Definite Integrals

A definite integral calculates the accumulation of a quantity, represented as the area under a curve, between two specified limits. In this case, the integral is evaluated from 0 to π/2, meaning we are interested in the area under the curve of the function sin(θ) / (1 + cos²(θ)) from θ = 0 to θ = π/2.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, relate angles to ratios of sides in right triangles. Understanding these functions is crucial for evaluating integrals involving them, as they often appear in calculus problems. The integral in question involves sin(θ) and cos²(θ), which are fundamental in trigonometric identities and integration techniques.

Integration Techniques

Integration techniques are methods used to find the integral of a function, which can include substitution, integration by parts, or recognizing standard forms. For the given integral, recognizing the structure of the integrand and possibly using a substitution can simplify the evaluation process, making it easier to compute the area under the curve.