trig substitution: int [ x^3 / sqrt[x^2 + 4] ] dx
- Thread starter cheffy
- Start date
You're down to:
\(\displaystyle \L\\8\int{tan^{3}(y)sec(y)}dy\)
Factor out tany
\(\displaystyle \L\\8\int{tan^{2}ytan(y)sec(y)}dy\)
\(\displaystyle tan^{2}(y)=sec^{2}(y)-1\)
\(\displaystyle \L\\8\int{(sec^{2}(y)-1)tan(y)sec(y)}dy\)
Let \(\displaystyle u=sec(y), \;\ du=sec(y)tan(y)dy\)
\(\displaystyle \L\\8\int{(u^{2}-1)}du\)
After you integrate, resub. If you want it in terms of x again,
Use \(\displaystyle sec(y)=\frac{\sqrt{x^{2}+4}}{2}\)
\(\displaystyle \L\\8\int{tan^{3}(y)sec(y)}dy\)
Factor out tany
\(\displaystyle \L\\8\int{tan^{2}ytan(y)sec(y)}dy\)
\(\displaystyle tan^{2}(y)=sec^{2}(y)-1\)
\(\displaystyle \L\\8\int{(sec^{2}(y)-1)tan(y)sec(y)}dy\)
Let \(\displaystyle u=sec(y), \;\ du=sec(y)tan(y)dy\)
\(\displaystyle \L\\8\int{(u^{2}-1)}du\)
After you integrate, resub. If you want it in terms of x again,
Use \(\displaystyle sec(y)=\frac{\sqrt{x^{2}+4}}{2}\)
- Joined
- Apr 12, 2005
- Messages
- 11,339
What? I would be rather surprised to find a textbook with this terminology. If we were to define a "correct method", I think it would be more like this, "It produces the desired result and the student understands it."Trebor said:
Is that a verb? There is no need to make communication even more difficult than it is already.
Here's why:
Derivation of the derivative of arctan(x):
\(\displaystyle y=arctan(x)\)
\(\displaystyle tan(y)=x\)
\(\displaystyle \frac{dy}{dx}sec^2(y)=1\)
\(\displaystyle \frac{dy}{dx}=\frac{1}{sec^2(y)}\)
\(\displaystyle 1+tan^2(y)=sec^2(y)\)
\(\displaystyle \frac{dy}{dx}=\frac{1}{1+tan^2(y)}=\frac{1}{1+x^2}\)
So you cannot substitute using y=arctan(x/2).
Derivation of the derivative of arsinh(x):
\(\displaystyle y=arsinh(x)\)
\(\displaystyle sinh(y)=x\)
\(\displaystyle \frac{dy}{dx}cosh(y)=1\)
\(\displaystyle \frac{dy}{dx}=\frac{1}{cosh(y)}\)
\(\displaystyle 1+sinh^2(y)=cosh^2(y)\)
\(\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{1+sinh^2(y)}}=\frac{1}{\sqrt{1+x^2}}\)
And that is why you need this hyperbolic substitution y = arsinh(x/2). And also why it is the correct method, as it obtains the correct answer.
Secondly, I can't see the problem with "square rooted". If you'd rather say "the denominator is raised to the power of a half", then by all means you do that. Personally, I don't think it's a problem.
Derivation of the derivative of arctan(x):
\(\displaystyle y=arctan(x)\)
\(\displaystyle tan(y)=x\)
\(\displaystyle \frac{dy}{dx}sec^2(y)=1\)
\(\displaystyle \frac{dy}{dx}=\frac{1}{sec^2(y)}\)
\(\displaystyle 1+tan^2(y)=sec^2(y)\)
\(\displaystyle \frac{dy}{dx}=\frac{1}{1+tan^2(y)}=\frac{1}{1+x^2}\)
So you cannot substitute using y=arctan(x/2).
Derivation of the derivative of arsinh(x):
\(\displaystyle y=arsinh(x)\)
\(\displaystyle sinh(y)=x\)
\(\displaystyle \frac{dy}{dx}cosh(y)=1\)
\(\displaystyle \frac{dy}{dx}=\frac{1}{cosh(y)}\)
\(\displaystyle 1+sinh^2(y)=cosh^2(y)\)
\(\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{1+sinh^2(y)}}=\frac{1}{\sqrt{1+x^2}}\)
And that is why you need this hyperbolic substitution y = arsinh(x/2). And also why it is the correct method, as it obtains the correct answer.
Secondly, I can't see the problem with "square rooted". If you'd rather say "the denominator is raised to the power of a half", then by all means you do that. Personally, I don't think it's a problem.
ChaoticLlama
Junior Member
- Joined
- Dec 11, 2004
- Messages
- 199
Try making the substitution u = Sqrt(x² + 4)
then... x² = u² - 4
and ... du = (x * dx)/Sqrt(x² + 4)
Therefore your intergral is |(u² - 4)du
=)
then... x² = u² - 4
and ... du = (x * dx)/Sqrt(x² + 4)
Therefore your intergral is |(u² - 4)du
=)
- Joined
- Apr 12, 2005
- Messages
- 11,339
I dare you to find it in a text book - even one. How about, "Well, spank me into Thursday! There's one o'them dang square roots thingys down thar, Fellers!" You are free to be as low with your language as you like in your personal papers. If you wish to communicate with the rest of the world, you may wish to upgrade just a bit. That's all I'm suggesting.Trebor said:
