Trig Substitution Problem

[Jason76]

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

Use x substitution, not u substitution. :idea:

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

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

\(\displaystyle \int \dfrac{3 \cos \theta}{\sqrt{9 - 9 \sin^{2} \theta}} d\theta\) Note: dx is replaced here :idea:

\(\displaystyle \int \dfrac{3 \cos \theta}{3 \cos \theta} d\theta\) What's going on here, with regards to the denominator?

\(\displaystyle \int 1 d\theta \rightarrow \theta + C\) Can't leave it this way, so must solve the \(\displaystyle x\) equation for \(\displaystyle \theta\).

You can draw a triangle, but not necessary.

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

\(\displaystyle \sin \theta = \dfrac{x}{3}\)

\(\displaystyle \theta = \arcsin(\dfrac{x}{3})\)

\(\displaystyle \int 1 d\theta \rightarrow \theta + C \rightarrow \arcsin(\dfrac{x}{3}) + C\) Back substitute into \(\displaystyle \theta\) to get final answer

 

[Deleted member 4993]

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

Use x substitution, not u substitution. :idea:

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

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

\(\displaystyle \int \dfrac{3 \cos \theta}{\sqrt{9 - 9 \sin \theta}} d\theta\) Note: dx is replaced here :idea:... careless mistake ... should be sin²(Θ) in the denominator

\(\displaystyle \int \dfrac{3 \cos \theta}{3 \cos \theta} d\theta\) ... recovered from mistake above

\(\displaystyle \int 1 d\theta \rightarrow \theta + C\) Can't leave it this way, so must solve the \(\displaystyle x\) equation for \(\displaystyle \theta\).

You can draw a triangle, but not necessary.

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

\(\displaystyle \sin \theta = \dfrac{x}{3}\)

\(\displaystyle \theta = \arcsin(\dfrac{x}{3})\)

\(\displaystyle \int 1 d\theta \rightarrow \theta + C \rightarrow \arcsin(\dfrac{x}{3}) + C\) Back substitute into \(\displaystyle \theta\) to get final answer

Good work .... as far as I can see - you have the correct answer.

 

[Jason76]

I have the right answer (well, actually copied it from a video), but I understand everything except that line I pointed out. I think it has something to do with \(\displaystyle 1 - \sin^{2}\theta = \cos^{2} \theta\)

 

[Deleted member 4993]

I have the right answer (well, actually copied it from a video), but I understand everything except that line I pointed out. I think it has something to do with \(\displaystyle 1 - \sin^{2}\theta = \cos^{2} \theta\)

Correct - derived from the Pythagorean theorem

 

[Jason76]

What does the Pythagorean theorem do in this case (for that line)?

I believe

\(\displaystyle \sqrt{9 - 9\sin^{2} \theta}\)

will give

\(\displaystyle 3 - 3\sin \theta\)

What step is next?

 

[daon2]

What does the Pythagorean theorem do in this case (for that line)?

I believe

\(\displaystyle \sqrt{9 - 9\sin^{2} \theta}\)

will give

\(\displaystyle 3 - 3\sin \theta\)

What step is next?

Since when does \(\displaystyle \sqrt{a-b}=\sqrt{a}-\sqrt{b}\)??

 

[Deleted member 4993]

What does the Pythagorean theorem do in this case (for that line)?

I believe

\(\displaystyle \sqrt{9 - 9\sin^{2} \theta}\)

will give

\(\displaystyle 3 - 3\sin \theta\)

What step is next?

Your belief is wrong.

Pythagorean theorem is used to prove

cos²(Θ) + sin²(Θ) = 1

which yields

1 - sin²(Θ) = cos²(Θ)

 

[Jason76]

This would be the missing step:

\(\displaystyle \sqrt{9 - 9\sin^{2}\theta} \rightarrow \sqrt{9(1 - \sin^{2}\theta)}\rightarrow \sqrt{9(\cos \theta)} \rightarrow 3 \cos \theta\)

 

[Jason76]

One thing to note is how u substitution can't be used, cause in the \(\displaystyle du = dx\) situation, there is no way to manipulate \(\displaystyle 2x\), so that it disappears from the right side of the \(\displaystyle du = dx\) equation.

Original equation has only \(\displaystyle dx\), not \(\displaystyle x dx\)

\(\displaystyle u = x^{2}\)

\(\displaystyle du = 2x dx\) - cannot manipulate, so must use trig substitution

 

[Deleted member 4993]

This would be the missing step:

\(\displaystyle \sqrt{9 - 9\sin^{2}\theta} \rightarrow \sqrt{9(1 - \sin^{2}\theta)}\rightarrow \sqrt{9(\cos \theta)} \rightarrow 3 \cos \theta\) ........ Incorrect

That should be:

\(\displaystyle \sqrt{9 - 9\sin^{2}\theta} \rightarrow \sqrt{9(1 - \sin^{2}\theta)}\rightarrow \sqrt{9(\cos^2 \theta)} \rightarrow 3 \cos \theta\)

 

[Jason76]

I noticed that right before your last post.