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Thread: Boring Calculus question (trigonometric substitution)

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    Default Boring Calculus question (trigonometric substitution)

    Can anyone explain the theory behind trigonometric substitution to me? The 'why' part? I can obviously input the identities, but I have no idea why I'm doing it or what it means.

    Thanks!

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    Default Re: Boring Calculus question (trigonometric substitution)

    I'm a little rusty (been a lot of years since Engine school), but based on my recollection and some Googling, here's the deal...

    Trigonometric substitution is just an extension of simple substitution, using trigonometric identities.

    Here's a nice page that talks about it: http://www.karlscalculus.org/calc11_4.html

    Here is a pretty good page on simple substitution: http://www.karlscalculus.org/calc11_2.html

    It seems that simple substitution is based on the chain rule. The chain rule says:

    if function f is differentiable at x and
    if function g is differentiable at f(x),
    then the composition h = g(f) is differentiable at x, and its derivative there is

    h'(x) = g'(f(x)) * f'(x)

    (I'm using * to represent the dot of multiplication.)

    Now, suppose you have an indefinite integral that looks like this (using S for the integral sign):

    S f(g(x))g'(x) dx

    If you write u = g(x), then du = g'(x) dx, and substituting gives you

    S f(g(x))g'(x) dx = S f(u) du

    This is the indefinite integral form of the chain rule.

    Integration by substitution woks whenever then integrand function is recognizable in the form f(g(x))g'(x) .

    In simple substitution, you were trying to find a function, u = g(x), that you could substitute in order to make the equation easier to solve. In trigonometric, you are instead pretending that x is a function of u: x = g(u), and substituting that function for x in the equation. if g(u) is a trigonometric function, you are transforming the equation to be a function of theta instead of a function of x. This is a nice thing because the integrals and derivatives of the trig functions are easy to find, and there are several well-known identities that help you rearrange your problem to be easier to solve. When you have solved the integral, you reverse the substitution you originally did, and you are back to a function of x.


    I hope this helps a bit. If you are still confused, and nobody steps up who is more up-to-date on calc than I am, feel free to ask more questions, and I will try to help.

    Be well...

    Lynn

    P.S. I really like your little picture. I fondly remember taking calc exams where I felt like that.

    L

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