91Ó°ÊÓ

Trigonometric integrals often involve products of sine and cosine functions. When faced with such an integral, a handy technique is to convert these products into sums or differences using product-to-sum identities. This transformation simplifies the integration process. In our exercise, we have the integral \( \int \sin x \sin 2x \, dx \). By applying the product-to-sum identity, we rewrite this integral as \( \int \frac{1}{2}(\cos (x-2x) - \cos (x+2x)) \, dx \). This identity is derived from combining the sine and cosine angle sum and difference formulas, which ultimately express a product of trigonometric functions as a combination of simpler trigonometric expressions.
Here’s a general form you might find useful:

• \( \sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)] \)
• \( \cos A \cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)] \)
• \( \sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)] \)

By recasting our integral into a more manageable form, we set the stage for straightforward integration.

The integration by substitution technique is akin to reverse chain rule and is used to simplify integrals by substituting variables. In our problem, after applying the product-to-sum identity, we need to integrate expressions like \( \int \cos(-x) \, dx \) and \( \int \cos(3x) \, dx \).
For \( \int \cos(3x) \, dx \), you substitute \( u = 3x \), which means \( du = 3 \, dx \), or \( dx = \frac{1}{3} \, du \). This substitution turns our integral into \( \int \cos(u) \cdot \frac{1}{3} \, du \), which simplifies to \( \frac{1}{3} \int \cos u \, du \).
The integral of \( \cos u \) is \( \sin u \), so we integrate and substitute back \( u = 3x \), resulting in \( \frac{1}{3} \sin(3x) \). Similarly, if you face \( \cos(-x) \), use the identity \( \cos(-x) = \cos x \), and integrate directly. Such substitutions simplify the process and make handling more complex expressions effective.

When we talk about definite integrals, we're discussing the integral over a specific interval. In our exercise, we tackled an indefinite integral, but the concepts extend naturally. A definite integral, represented by \( \int_{a}^{b} f(x) \, dx \), provides the area under the curve of \( f(x) \) from \( x = a \) to \( x = b \). Such integrals yield a number, representing the net area between the function and the x-axis, taking into account the parts below the x-axis as negative.
Using the Fundamental Theorem of Calculus, we can evaluate these by finding an antiderivative \( F(x) \) of \( f(x) \). Then, compute \( F(b) - F(a) \). This method is crucial in physics and engineering for analyzing continuous systems over time or certain intervals. While our exercise focused on an indefinite integral, the step-by-step breakdown and simplifications would be similar even if boundaries were given.