Trig Function Substitution

[Jason76]

Where \(\displaystyle a = 9\) and \(\displaystyle x = a\sin\theta\)

Formula: \(\displaystyle \sqrt{a^{2} - x^{2}}\)

\(\displaystyle \int \dfrac{1}{9 - x^{2}} dx\)

\(\displaystyle \int \dfrac{1}{\sqrt{9 - 9\sin^{2}\theta }}dx\)

\(\displaystyle \int \dfrac{1}{\sqrt{9(1 - \sin^{2}\theta)}}dx\)

\(\displaystyle \int \dfrac{1}{\sqrt{9(\cos^{2}\theta)}}dx\)

\(\displaystyle \int \dfrac{1}{3\cos\theta}dx\)

Using differentiation:

\(\displaystyle u = 3\sin\theta\)

\(\displaystyle du = 3\cos\theta\)

\(\displaystyle \int \dfrac{1}{3\cos\theta}(3\cos\theta)dx\)

\(\displaystyle \int \dfrac{3\cos\theta}{3\cos\theta} dx\)

\(\displaystyle \int 1 dx = \theta + C\)

Does this look right? Am I using the right chain of thought? But the next step would be to find \(\displaystyle \theta\) cause we can't have this final answer.

 

[lookagain]

Where \(\displaystyle a = 9\) and \(\displaystyle x = a\sin\theta\)

Formula: \(\displaystyle \sqrt{a^{2} - x^{2}}\)

\(\displaystyle \int \dfrac{1}{9 - x^{2}} dx\) . . . You are missing the square root symbol around the denominator.

\(\displaystyle \int \dfrac{1}{\sqrt{9 - 9\sin^{2}\theta }}dx\)

\(\displaystyle \int \dfrac{1}{\sqrt{9(1 - \sin^{2}\theta)}}dx\)

\(\displaystyle \int \dfrac{1}{\sqrt{9(\cos^{2}\theta)}}dx\)

\(\displaystyle \int \dfrac{1}{3\cos\theta}dx\)

Using differentiation:

\(\displaystyle u = 3\sin\theta\) . . . No, x = \(\displaystyle 3\sin\theta\)

\(\displaystyle du = 3\cos\theta\) . . . Don't forget the \(\displaystyle d \theta.\) . . . No, it's supposed to be "dx ="

\(\displaystyle \int \dfrac{1}{3\cos\theta}(3\cos\theta)d\theta\) . . . No, it is not "dx."

\(\displaystyle \int \dfrac{3\cos\theta}{3\cos\theta} d\theta\) . . . Same error as above.

\(\displaystyle \int 1 dx . . . Again, the same error. It should be \(\displaystyle \int 1 d\theta.\)

= \(\displaystyle \theta + C\)


You are not finished. You need to back-substitute to get it in terms of x.

\)

\(\displaystyle .\)

 

[Jason76]

Where \(\displaystyle a = 9\) and \(\displaystyle x = a\sin\theta\)

Formula: \(\displaystyle \sqrt{a^{2} - x^{2}}\)

\(\displaystyle \int \dfrac{1}{\sqrt{9 - x^{2}}} dx\)

\(\displaystyle \int \dfrac{1}{\sqrt{9 - 9\sin^{2}\theta }}dx\)

\(\displaystyle \int \dfrac{1}{\sqrt{9(1 - \sin^{2}\theta)}}dx\)

\(\displaystyle \int \dfrac{1}{\sqrt{9(\cos^{2}\theta)}}dx\)

\(\displaystyle \int \dfrac{1}{3\cos\theta}dx\)

Ok, let's stop here and go step by step. What is the next step (hint) and why?

Using differentiation:

\(\displaystyle x = 3\sin\theta\)

\(\displaystyle dx = 3\cos\theta\)

 

[Deleted member 4993]

Where \(\displaystyle a = 9\) and \(\displaystyle x = a\sin\theta\) ← That should be a = 3

dx = 3cos(Θ) dΘ

Formula: \(\displaystyle \sqrt{a^{2} - x^{2}}\)

\(\displaystyle \int \dfrac{1}{9 - x^{2}} dx\)

\(\displaystyle \int \dfrac{1}{\sqrt{9 - 9\sin^{2}\theta }}dx\) ← That should be \(\displaystyle \displaystyle \int \frac{1}{\sqrt{9 - 9\sin^{2}\theta }}3cos(Θ) dΘ\)

\(\displaystyle \int \dfrac{1}{\sqrt{9(1 - \sin^{2}\theta)}}dx\)

\(\displaystyle \int \dfrac{1}{\sqrt{9(\cos^{2}\theta)}}dx\)

\(\displaystyle \int \dfrac{1}{3\cos\theta}dx\)

= \(\displaystyle \displaystyle \frac{1}{3}\int sec(\theta) *3*cos(\theta)d\theta\)

= \(\displaystyle \displaystyle \int d\theta\)

= Θ + C

= sin^-1(x/3) + C



Using differentiation:

\(\displaystyle u = 3\sin\theta\)

\(\displaystyle du = 3\cos\theta\)

\(\displaystyle \int \dfrac{1}{3\cos\theta}(3\cos\theta)dx\)

\(\displaystyle \int \dfrac{3\cos\theta}{3\cos\theta} dx\)

\(\displaystyle \int 1 dx = \theta + C\)

Does this look right? Am I using the right chain of thought? But the next step would be to find \(\displaystyle \theta\) cause we can't have this final answer.

.

 

[Jason76]

.

Thanks for the help. There seems to be a couple of different ways which all came up with the same answer. I was also sent a private message with another way, as well as the way shown on the video I was watching. The guy on the video proceeded to solve for \(\displaystyle x\), rather than accept an \(\displaystyle \arcsin\) answer. He did so by making a triangle and putting \(\displaystyle x\) on on side, \(\displaystyle 3\) on the other, then then \(\displaystyle \sqrt{9 - x^{2}}\) on the other, and then solving for \(\displaystyle \theta\). But in the end, came up with the same answer of everybody else (an \(\displaystyle \arcsin\) variation).

 

[Deleted member 4993]

You are making careless mistakes - like not replacing dx with f'(Θ)dΘ while substituting x = f(Θ). That created most of the confusions in the problem above. This can become really a problem when you try to advanced level of calculus (like definite integrals, integrals with multiple variables, etc.).

 

[Jason76]

You are making careless mistakes - like not replacing dx with f'(Θ)dΘ while substituting x = f(Θ). That created most of the confusions in the problem above. This can become really a problem when you try to advanced level of calculus (like definite integrals, integrals with multiple variables, etc.).

Right, I need more experience. Will work on it.

 

[Jason76]

Actually, you don't need trig or u substitution to solve the problem.

You got: \(\displaystyle \int \dfrac{du }{a^{2} - u^{2}} = \sin^{-1} \dfrac{u}{a} + C\) Nonetheless, solving it by trig substitution (and doing a triangle) is a good way to PROVE it. Well, for one thing, you can automatically tell from the "get go" what \(\displaystyle a\) amd \(\displaystyle x{/tex] are. There is no need to go any further. You just implement the formula.\)