91Ó°ÊÓ

When faced with complex integrals, especially those involving compositions of functions, the substitution method is a powerful tool. It's like a reverse application of the chain rule used in differentiation. In the process, we introduce a new variable, often labeled as \( u \), to simplify the integral into a more manageable form.

In this exercise, we began with the function \( f(x) = \sin(2 - 5x) \). By setting \( u = 2 - 5x \), the integral becomes less complex. Differentiating \( u \) with respect to \( x \) gives \( du/dx = -5 \), or rearranging, \( dx = du/-5 \). The original integral is transformed into \( \int \sin(u) \cdot \left( -\frac{1}{5} \right) \, du \).

This approach makes integration simpler, turning a tangled expression into something that is straightforward to manage. Remember to replace \( u \) with the original variable at the end to ensure the antiderivative is expressed in terms of the original variable.

The chain rule in calculus lets us differentiate composite functions efficiently. It's harnessed in both differentiation and integration. While differentiation uses the chain rule directly, integration employs a form of it through substitution.

In the solution, after integrating directly with substitution, we used the chain rule again to verify our result. The derived antiderivative is \( F(x) = \frac{1}{5} \cos(2 - 5x) + C \). When differentiating \( F(x) \) to check our work, the chain rule applies: we multiply the derivative of the outer function, \( \cos(u) \), by the derivative of the inner function, \(-5x \). Thus, the derivative \( F'(x) = \frac{1}{5} \cdot (-\sin(2 - 5x)) \cdot (-5) \) conveniently returns to our original function \( \sin(2 - 5x) \), confirming our work.

This highlights the connection between the differentiation and integration processes through the chain rule.

Trigonometric integration deals with integrating functions involving trigonometric functions like sine and cosine. It's a common part of calculus as many periodic phenomena are modeled using these functions.

In this exercise, the key challenge was integrating \( \sin(u) \). The derivative of cosine is negative sine, so integrating \( \sin(u) \) yields \( -\cos(u) \). Thus, the trigonometric integral \( \int \sin(u) \, du \) results in \( -\cos(u) \). After applying substitution, it simplifies the integration of complex trigonometric functions by transforming them into basic integrals.

Remembering these basic trigonometric integrals speeds up solving problems involving sine and cosine, making it easier to analyze and interpret the results efficiently.