Summary
Highlights
The session begins by introducing inverse trigonometric and hyperbolic functions. The instructor emphasizes understanding how to solve problems using these functions rather than dwelling on deep theoretical explanations. The main goal is to equip students with sufficient integration techniques to solve various problems, which often involves memorizing specific formulas for different function types.
The instructor demonstrates solving an integral problem (integral of dx / 2 + 9x²) that requires rewriting the expression to fit an inverse tangent formula. This involves factoring out constants and identifying 'u' and 'a' terms. The solution leads to 1/(3√2) arctan(3x/√2) + C.
Another example (integral of dx / x * √(4x² - 9)) is tackled, which fits the arc secant formula. The process involves factoring constants from under the square root and identifying 'u' and 'a' values. The final answer is 1/3 arcsec(|2x|/3) + C.
A more complex problem (integral of dx / (x² - 4x + 7)) is presented, where the denominator is a trinomial. The strategy is to complete the square to transform the denominator into the form a² + u², allowing the application of the arctan formula. The final result is 1/√3 arctan((x-2)/√3) + C.
The lesson transitions to hyperbolic functions (sinh, cosh, etc.) and their integrals. The focus remains on applying formulas. An example (integral of cosh(2x) sinh²(2x) dx) demonstrates using u-substitution twice to simplify the integral into a basic power rule, resulting in sinh³(2x)/6 + C.
Integration by parts is introduced as the integration equivalent of the product rule for derivatives. The formula ∫ u dv = uv - ∫ v du is presented. A key strategy for choosing 'u' and 'dv' is highlighted using the 'LIATE' (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) priority rule.
The first example for integration by parts is integral(x e^x dx). Following the LIATE rule, u is set to 'x' (algebraic) and dv to 'e^x dx' (exponential). After applying the formula, the solution is x e^x - e^x + C.
The integral(x² ln x dx) is solved using integration by parts. 'ln x' (logarithmic) is chosen as 'u' due to LIATE priority. The process involves differentiating 'u' and integrating 'dv' (x² dx), leading to a solution of (x³ ln x)/3 - x³/9 + C.
An example (integral(x² sin x dx)) requiring multiple applications of integration by parts is demonstrated. Choosing u = x² (algebraic) and dv = sin x dx (trigonometric) initially. The resulting integral (integral(x cos x dx)) also needs integration by parts. The final answer is -x² cos x + 2x sin x + 2 cos x + C.
The session concludes with trigonometric integrals involving products of sine and cosine functions with different arguments (e.g., sin(mx)cos(nx)). Specific product-to-sum formulas are used to rewrite these integrals into simpler forms that can be integrated directly. An example of integral(sin(5x)cos(4x) dx) results in -cos(x)/2 - cos(9x)/18 + C.