Intro to Trigonometric Substitution --- Ex: Deriving Area of Circle Formula

Trigonometry is great for integration because we can utylize all the various trigonometric identities to manipulate challenging integrals into easier ones. Trigonometric SUBSTITUTION involves making substitutions like x=sin(theta) to introduce trig, and then use the trig identities.

In this first video we will look at the definite case and see an example that arises naturally: determining the area of a circle formula. There will be a key technical detail about restricting the domain of the substitution to ensure the function is 1:1.

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This video was created by Dr. Trefor Bazett, an Assistant Professor, Educator at the University of Cincinnati. #calculus #math

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[tempritempri]

I just watched two famous instructors explaining trigonometric substitution, and this guy surpassed them exponentially 💯‼️😇

[nathankraft]

I really wish I had you as my actual professor.

[aydyncoatlbush]

You are my go to every week when i don't understand CALC 2

[kristoffercorbyn]

thanks, I got a homework problem like this in my calc 1 class and they wanted us to come up with the area by graphing, it really irked me because when I tried to solve the integral given what I have been taught so far about evaluating integrals, I couldn't come to an answer that made sense.

[viktorashistan]

0:55 mans edited him over with different volume, jk almost unnoticeable if you don't look for it

[naiko]

If someone's confused, careful here!
We are basically doing u-substitution in the opposite way! It's as if x was already our u, and we wanted to replace u with a function! (usually we go the other way around)
Since when we substitute a function with u we also take away it's derivative from the integral, this time we have to put it back in.

[duckymomo]

Is trig substitution called change of Variables in general case?

A change to polar coordinates?

(I’m trying to abstract it to vector calculus, real analysis)

[emperoy]

cosθ = x/r sinθ = y/r. We should know that sinθ=opposite/hypotenuse and cosθ=adjacent/hypothenuse. so, x=rcosθ and y=rsinθ; Can you justify your operation? Thanks