Take Derivatives of Inverse Trig Functions (ArcSin, ArcCos) - [2]

This section of the 'Advanced Calculus 2' tutorial focuses on the derivatives and integrals of inverse trigonometric functions. The instructor plans to define the derivatives of each inverse trig function and work through problems involving them. The session builds upon the basics of inverse functions covered in the previous section, including their existence and range of angles. The derivatives of inverse sine, cosine, tangent, cotangent, secant, and cosecant functions are introduced as formulas to be used rather than derived in detail, emphasizing application over proof. The instructor also explains that knowing these derivatives inherently defines the integrals of certain functions, allowing students to solve more complex problems.

The tutorial delves into the specifics of finding derivatives of functions involving inverse trigonometric operations. The instructor provides the derivative formulas for inverse sine, cosine, tangent, cotangent, secant, and cosecant functions, highlighting the symmetry in their forms. For instance, the derivative of the inverse sine function is given as 1 over the square root of (1 - x^2), while the inverse cosine derivative is the negative of that. The inverse tangent and cotangent derivatives are also presented, showcasing a similar pattern. The importance of the chosen ranges for these functions in creating these symmetrical and simple derivative forms is discussed. The session also covers how to use these derivatives to solve integral problems, with the integral of the derivative of an inverse function naturally leading back to the original inverse function plus a constant.

This part of the script illustrates how to apply the derivatives of inverse trigonometric functions to solve calculus problems. The instructor uses the chain rule to find the derivatives of composite functions, such as the inverse tangent of x cubed and the inverse sine of x squared. The process involves taking the derivative of the outer function and then multiplying by the derivative of the inner function, as demonstrated through step-by-step calculations. The examples given help students understand how to manipulate and simplify expressions involving inverse trigonometric functions to find their derivatives, preparing them for solving problems on tests and further studies.

The script moves on to integrals, starting with a definite integral involving the inverse tangent function. The instructor demonstrates how to evaluate the integral by recognizing the form that matches the integral of the derivative of the inverse tangent function. The process involves pulling out constants and evaluating the inverse tangent function at the limits of integration. The importance of knowing the range of angles for the tangent function is emphasized, and the evaluation of the integral results in an expression involving the difference of two inverse tangent functions. This section highlights the connection between derivatives and integrals and the practical application of inverse trigonometric functions in integral calculus.

The instructor introduces a technique for solving more complex integrals involving inverse trigonometric functions by using substitution. An example is given where the integral of x squared over the square root of (1 - x to the sixth power) is simplified by letting u be x to the third power. This substitution transforms the integral into a simpler form that resembles the integral of the derivative of the inverse sine function. The process involves taking the derivative of u with respect to x, solving for dx, and then substituting back into the integral. The result is an expression involving the inverse sine of u, plus a constant, which is then evaluated in terms of the original variable x. This method showcases a strategic approach to tackling integrals that can be related to inverse trigonometric functions.