Evaluating integrals Evaluate the following integrals.

∫π/₆^π/³ (sec² t + csc² t) dt

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Step 1: Recognize that the integral involves the sum of two trigonometric functions: sec²(t) and csc²(t). Recall their respective antiderivatives: ∫sec²(t) dt = tan(t) and ∫csc²(t) dt = -cot(t).

Step 2: Break the integral into two separate integrals: ∫(sec²(t) + csc²(t)) dt = ∫sec²(t) dt + ∫csc²(t) dt.

Step 3: Compute the antiderivative of each term. For sec²(t), the antiderivative is tan(t). For csc²(t), the antiderivative is -cot(t). Combine these results to get tan(t) - cot(t).

Step 4: Apply the Fundamental Theorem of Calculus to evaluate the definite integral. Substitute the upper limit π/3 and the lower limit π/6 into the antiderivative expression tan(t) - cot(t).

Step 5: Simplify the result by calculating tan(π/3), tan(π/6), cot(π/3), and cot(π/6). Subtract the values obtained at the lower limit from those at the upper limit to find the final value of the integral.

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Integration

Integration is a fundamental concept in calculus that involves finding the accumulated area under a curve represented by a function. It is the reverse process of differentiation and is used to calculate quantities such as areas, volumes, and total accumulated change. The integral symbol (∫) denotes the operation, and definite integrals, like the one in the question, have specified limits of integration.

Trigonometric Functions

Trigonometric functions, such as secant (sec) and cosecant (csc), are essential in calculus, particularly in integrals involving angles. The secant function is defined as the reciprocal of the cosine function, while the cosecant function is the reciprocal of the sine function. Understanding their properties and relationships is crucial for evaluating integrals that include these functions.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation with integration, stating that if a function is continuous on an interval, then the integral of its derivative over that interval equals the change in the function's values at the endpoints. This theorem allows us to evaluate definite integrals by finding an antiderivative of the integrand and applying the limits of integration, simplifying the process of calculating areas under curves.

Evaluating integrals Evaluate the following integrals.∫π/₆^π/³ (sec² t + csc² t) dt