91Ó°ÊÓ

Definite integrals are a fundamental concept in calculus used to calculate the area under a curve within a specific interval. In our problem, the definite integral is given as \(\textstyle \int_{0}^{\frac{5 \pi}{6}} 3x \cos x dx\). Unlike indefinite integrals, definite integrals have upper and lower limits, which are 0 and \(\textstyle \frac{5 \pi}{6}\) in this case. The evaluation of a definite integral results in a numerical value that represents the accumulated area between the curve and the x-axis over the given interval. We start by applying the integration by parts formula and periodically integrating and differentiating until we simplify and evaluate the integral using the provided limits.

Trigonometric integrals involve the integration of combinations of trigonometric functions like sine and cosine. These functions have specific properties and integrals that make them manageable. Let's dissect the trigonometric part of our original problem: the integral \(\textstyle \int_{0}^{\frac{5 \pi}{6}} 3x \cos x dx\). Here, we are dealing with the cosine function. During the process, we use properties like \(\textstyle \sin(x)\) and known values, such as \(\textstyle \sin(\frac{5 \pi}{6}) = \frac{1}{2}\) and \(\textstyle \cos(\frac{5 \pi}{6}) = -\frac{ \sqrt{3}}{2}\), to help simplify the evaluation. Understanding these basic trigonometric functions and their integrals is essential for breaking down more complex integrations.

Calculus is rich with techniques to solve integrals, such as the Integration by Parts method we used for our problem. Integration by parts is derived from the product rule for differentiation. The formula we use is \(\textstyle \int u dv = uv - \int v du\). Choosing \(\textstyle u = 3x\) and \(\textstyle dv = \cos x dx\) allowed us to break the original integral into more manageable pieces. We then differentiated \(\textstyle u\) to get \(\textstyle du = 3 dx\) and integrated \(\textstyle dv\) to get \(\textstyle v = \sin x\). Plugging these into the formula gave us a simpler integral to solve. Mastering such techniques allows one to tackle a broader range of integrals effectively.

Proper integration involves a series of methodical steps. In our example, we started by identifying a suitable method: Integration by Parts. We chose the functions for \(\textstyle u\) and \(\textstyle dv\), then calculated their derivatives and integrals respectively. After applying the formula, we evaluated the boundary terms by substituting the integral limits. This required substituting back each term, simplifying the trigonometric expressions, and summing them up. Finally, combining all simplified parts provided the final answer of the definite integral. This structured approach ensures that complex integrals are broken down into more straightforward computations.