Antidifferentiating the Inverses of Trigonometric Functions

OVERVIEW:

• Formulas for the derivatives and the antiderivatives of the inverses of the trigonometric functions
• Antidifferentiating the inverses of trigonometric functions using techniques such as completing the square, the chain rule, substitution, and multiplying by one and using the constant multiple rule
• Proving an antiderivative of a derivative is equal to the derivative of an antiderivative

Slide 2

These are the derivative and antiderivative rules that we discovered in the previous class.


Slide 3

The denominator with a square root! Factoring won't do us any good here (I've tried already) because we want the fraction to look similar to arcsinx, so we complete the square. *zeph thinks back to grade 11 precal*

AND VOILA! We were lucky. It looks very similar to the arctanx function when, using the method of substitution, we let u = x - 3 and find the derivative of u (du), and also because the denomiator has 1 - u as its radicand. (But what if that 1 was another constant? We'll look at that later, but for now, let's solve this problem first!) Using the antderivative rules, we know that the antiderivative of 1/sqrt(1-u^2) is arcsinu + C. Resubsituting u into arcsinu + C, we get the answer!


Slide 4

Now what if that 1-(x-3)^2 underneath the radicand was 5-(x-3)^2? There's no 5 in the derivative of arcsinx, so we put it there by doing a neat trick. In the green on the right side of the slide, we factored out a five, rewrote (x-3)^2/5 as ((x-3)/sqrt(5))^2 (since 5 = sqrt(5)^2), and rewrote (x-3)/sqrt(5) as x/sqrt(5) - 3/sqrt(5) in blue. We let the blue stuff equal u and differentiated it to du. Rewriting the expression in terms of u, we get 1/sqrt(1-u^2) which we can resubstitute u back in to get the answer.


Slide 5

To let the left-hand side equal the right-hand side, let's take the derivative of both sides since we're dealing with antiderivatives. Differentiating arcsin(x/a), via chain rule, and replacing the resulting denominator with an algebraic equivalent (look in box--we let the 1 = a^2/a^2 and factored out 1/a^2) to get 1/sqrt(a^2-x^2). Therefore, left-hand side equals right-hand side.

Slide 6

For the left question, we used what we generalized in the box in slide 5 to "simplify" the denominator. Then we factored out a 1/10 using the constant multiple rule. Using the method of substitution, we let u = 3x/10, find its derivative (du), and rewrite the expression in terms of u. We realize that the antiderivative of the blue is arcsinu+C. We can then resubstitute u back into the expression to get the answer.

The middle and right questions we haven't done yet.


HOUSEKEEPING:

• Breaking the chain, Hi I'm Justus, you're up next.
• Middle and right questions, do for homework, I assume.
• Today is Day 3 of the school day cycle. Today is an optional class.
  
• All calculus students will be writing the grade 12 contest on the 18th, which are able to win $1000, but .:. J + ME .:. and Hi I'm Justus may need to reschedule their plans for the DMCI Jr. High Tours on the 18th and 19th.
• You may achieve a high school credit for tutoring an EAL student in math for 100 hours.
• And don't forget to listen to the Dr. Love messages at the last ten minutes of class!