Evaluating integrals Evaluate the following integrals.

∫ (cos 7ω) /(16 + sin² 7ω) dω

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Step 1: Recognize that the integral involves a trigonometric function in the numerator and a combination of trigonometric functions in the denominator. This suggests that substitution might simplify the integral.

Step 2: Let u = sin(7ω). Then, compute the derivative of u with respect to ω: du/dω = 7cos(7ω), or equivalently, du = 7cos(7ω)dω.

Step 3: Rewrite the integral in terms of u. Substitute sin(7ω) with u and cos(7ω)dω with du/7. The integral becomes ∫ (1 / (16 + u²)) * (1/7) du.

Step 4: Factor out the constant 1/7 from the integral. The integral now simplifies to (1/7) ∫ (1 / (16 + u²)) du.

Step 5: Recognize that the integral ∫ (1 / (a² + u²)) du is a standard form, which evaluates to (1/a) * arctan(u/a) + C. Here, a² = 16, so a = 4. Apply this formula to complete the integration.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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 can be used to calculate quantities such as areas, volumes, and total accumulated change. The integral can be definite, providing a numerical value over a specific interval, or indefinite, resulting in a general form of antiderivatives.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are periodic functions that relate angles to ratios of sides in right triangles. In calculus, these functions are essential for modeling oscillatory behavior and are frequently encountered in integrals. Understanding their properties, such as periodicity and symmetry, is crucial for evaluating integrals involving trigonometric expressions.

Substitution Method

The substitution method is a technique used in integration to simplify the process of finding an integral. It involves substituting a part of the integrand with a new variable, which can make the integral easier to evaluate. This method is particularly useful when dealing with composite functions or when the integrand contains a function and its derivative, allowing for a more straightforward integration process.

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