Troubleshooting a Trigonometric Integral: Algebra and Solutions

In summary: $\int \frac{x^{3/2}}{\sqrt{4-x}}\,dx = \int \frac{x^2}{\sqrt{4-x}}\cdot\frac{dx}{\sqrt x} = \int \frac{(2\sin\theta)^4}{\sqrt{4-(2\sin\theta)^2}}\cdot 4\cos\theta\,d\theta$
  • #1
Kristal1
2
0
I have a few questions and a request for an explanation.

I worked this problem for a quite a while last night. I posted it here.

https://math.stackexchange.com/questions/3547225/help-with-trig-sub-integral/3547229#3547229

The original problem is in the top left. Sorry that the negative sometimes gets cut off in the photo, and yes I know it's not fully simplified there.

My first question is the more involved one: Is the algebra in my original work sound? If it is, why doesn't it work in this instance?

My second question is: is this a correct solution? https://imgur.com/2tgEz0O

It is for webwork, and I used two out of three chances. I'd prefer to keep my perfect webwork mark, obviously ;p

Finally, I was kind of impressed with Ans4's square completion and had to run it through to see that it was correct. That's such a useful skill. Do you have some advice about how I could improve my math tricks to that point?
 
Physics news on Phys.org
  • #2
Is the algebra in my original work sound?

Looks like you did this (correct me if I'm mistaken) ...

\(\displaystyle -7\int \dfrac{x^2}{\sqrt{4x-x^2}} \,dx = -7\int \dfrac{x^2}{\sqrt{x}\sqrt{4-x}} \,dx = -7\int \dfrac{x^{3/2}}{\sqrt{4-x}} \,dx\)

from here, I see you attempted to use the trig sub $\sqrt{x} = 2\sin{\theta}$.

you made a mistake determining $dx$ ...

$\sqrt{x} = 2\sin{\theta} \implies x = 4\sin^2{\theta} \implies dx = 8\sin{\theta}\cos{\theta} \, d\theta$
 
  • #3
I did sub it in as a replacement for $\sqrt{x}$ though.

I posted this elsewhere since I wasn't even sure if it would be approved here. There it was suggested that I should have put $4cos\theta=\frac{1}{\sqrt{x}}dx$

This is an important conceptual point, though. Are you saying that this is the only correct derivative because dx means wrt x and not x to some power?

If so, well it would be nice if a teacher told us these things sometimes...
 
  • #4
Kristal said:
I did sub it in as a replacement for $\sqrt{x}$ though.

I posted this elsewhere since I wasn't even sure if it would be approved here. There it was suggested that I should have put $4cos\theta=\frac{1}{\sqrt{x}}dx$

This is an important conceptual point, though. Are you saying that this is the only correct derivative because dx means wrt x and not x to some power?

If so, well it would be nice if a teacher told us these things sometimes...

Hi Kristal, welcome to MHB! ;)

There are 2 ways that are generally taught to do a substitution.

You have the integral
$$\int \frac{x^{3/2}}{\sqrt{4-x}}\,dx$$
and you want to replace $\sqrt x$ by $2\sin\theta$.

Method 1
Write $x$ as a function of $\theta$ and take the derivative as skeeter suggested:
$$\sqrt x=2\sin\theta \implies x=4\sin^2\theta \implies dx=d(4\sin^2\theta) = 8\sin\theta\cos\theta\,d\theta$$
We can now replace $\sqrt x$ and $dx$ to find:
$$\int \frac{x^{3/2}}{\sqrt{4-x}}\,dx = \int \frac{(2\sin\theta)^{3}}{\sqrt{4-(2\sin\theta)^2}}\,d(4\sin^2\theta)
= \int \frac{(2\sin\theta)^{3}}{\sqrt{4-(2\sin\theta)^2}}\cdot 8\sin\theta\cos\theta\,d\theta$$

Method 2
Take the derivative left and right, and do what needs to be done to eliminate $x$.
In this case:
$$\sqrt x=2\sin\theta \implies d(\sqrt x)=d(2\sin\theta) \implies \frac 1{2\sqrt x}\,dx= 2\cos\theta\,d\theta
\implies \frac{dx}{\sqrt x} = 4\cos\theta\,d\theta$$
Rewrite the integral a bit to match and substitute:
$$\int \frac{x^{3/2}}{\sqrt{4-x}}\,dx = \int \frac{x^2}{\sqrt{4-x}}\cdot\frac{dx}{\sqrt x}
= \int \frac{(2\sin\theta)^4}{\sqrt{4-(2\sin\theta)^2}}\cdot 4\cos\theta\,d\theta
$$

We can see that the result is the same in both cases.

I'm afraid I can't comment on what your teacher is telling you since I don't know what that is.
We can only point out and show what the correct math is.
 
  • #5
my calculator’s CAS worked out these antiderivatives ...
 

Attachments

  • 50048641-AAC5-4E4C-8850-48E9C774DBFC.jpeg
    50048641-AAC5-4E4C-8850-48E9C774DBFC.jpeg
    36.4 KB · Views: 81

FAQ: Troubleshooting a Trigonometric Integral: Algebra and Solutions

What is a trigonometric integral?

A trigonometric integral is an integral that involves trigonometric functions such as sine, cosine, tangent, etc. It is used to find the area under a curve of a trigonometric function or to solve equations involving trigonometric functions.

What are the common challenges in solving trigonometric integrals?

Some common challenges in solving trigonometric integrals include identifying the appropriate trigonometric substitution, dealing with complex trigonometric identities, and integrating trigonometric functions with multiple variables.

How do I know when to use trigonometric substitution?

Trigonometric substitution is typically used when the integral involves a radical expression or when the integral contains a trigonometric function raised to a power. It is also used when the integral involves the product of a trigonometric function and a polynomial.

What are some common techniques for solving trigonometric integrals?

Some common techniques for solving trigonometric integrals include using trigonometric identities, using trigonometric substitution, and using integration by parts. It is also helpful to simplify the integral by factoring out common terms or using trigonometric identities to rewrite the integral in a more manageable form.

Are there any tips for solving trigonometric integrals more efficiently?

Yes, some tips for solving trigonometric integrals more efficiently include practicing regularly, familiarizing yourself with common trigonometric identities, and breaking the integral down into smaller, more manageable parts. It is also helpful to double check your work and use a graphing calculator to verify your answer.

Similar threads

Replies
4
Views
2K
Replies
6
Views
2K
Replies
5
Views
2K
Replies
6
Views
658
Replies
1
Views
2K
Replies
1
Views
1K
Back
Top