Integral: Find x=sin θ for 0-0.5

[songoku]

Homework Statement

Find using substitution x=\sin \theta

\int _0^{0.5}\frac{x}{\sqrt{1-x^2}}dx

Homework Equations

integration

The Attempt at a Solution

\int _0^{0.5}\frac{x}{\sqrt{1-x^2}}dx

=\int_0^{\frac{\pi}{6}}\sin \theta\;d\theta

=1-\frac{1}{2}\sqrt{3}

What I want to ask is about changing the upper bound of the integral.

For x = 0.5 :
0.5 = \sin \theta

Here I choose \theta = \frac{\pi}{6}. But what if I choose \theta = \frac{5}{6}\pi\;??

So, the integral :
=\int_0^{\frac{5}{6}\pi}\sin \theta\;d\theta

=1+\frac{1}{2}\sqrt{3}

Thanks

 

[espen180]

A good way to check your answers on integration problems is to solve them in different ways and compare their answers. One way to do this is to use another integration method.

 

[arildno]

Be careful!

We have, ignoring boundary values:
\int\frac{xdx}{\sqrt{1-x^{2}}}=\int\frac{\sin\theta\cos\theta{d\theta}}{\sqrt{1-\sin^{2}\theta}}=\int\sin\theta\frac{\cos\theta}{|\cos\theta|}d\theta

Do you see what you need to be careful about when simplifying this further?

In particular, why is your choice: "But what if I choose \theta=\frac{5}{6}\pi" troublesome?

 

[songoku]

Hi arildno and espen180

We have, ignoring boundary values:
\int\frac{xdx}{\sqrt{1-x^{2}}}=\int\frac{\sin\theta\cos\theta{d\theta}}{\sqrt{1-\sin^{2}\theta}}=\int\sin\theta\frac{\cos\theta}{|\cos\theta|}d\theta

Do you see what you need to be careful about when simplifying this further?

In particular, why is your choice: "But what if I choose \theta=\frac{5}{6}\pi" troublesome?[/QUOTE]

Ah, maybe you are referring to the square root. The value of \cos \theta should be positive so the angle should be in first quadrant, where the values of sin and cos are positive.

A good way to check your answers on integration problems is to solve them in different ways and compare their answers. One way to do this is to use another integration method.

I've checked it using substitution u = 1-x^2 and got the answer. What I am confused is about why I can't take the upper bound to be (5 pi)/6. But I think I get it now

Thank you very much, arildno and espen180

 

[njama]

It is same either if you pick п/6 or 5п/6 because п - п/6 = 5п/6 and both result with positive value of sin(x).

What arildno meant to say is that you should be careful with the absolute value \int{sin(x)\frac{cos(x)}{|cos(x)|}dx}=\int{sin(x)\frac{cos(x)}{\pm cos(x)}dx}=\int{\pm sin(x)dx}

So, you should consider the negative values for cos(x) and not just the positive ones.

P.S And why you chose x=0 for sin(x)=0, why you did not choose x=п, since both are valid?

 

[songoku]

Hi njama

I don't really understand about the absolute sign. Why can it be negative? I think it's should be positive since it's a square root and we don't have to consider the negative value.

Thanks

 

[njama]

Ok, I will explain.

For example if you have y^2=16 so that we can write |y|=4

What that means is y can be either -4 or 4 so that |y|=4 or y=\pm 4

Lets check it out:

(-4)²=16, 4²=16 or |-4|=4 but also |4|=4

In this case take cos(x) = y, and you will see what I am talking about.

 

[HallsofIvy]

Hi njama

I don't really understand about the absolute sign. Why can it be negative? I think it's should be positive since it's a square root and we don't have to consider the negative value.

Thanks

Of course, |cos(\theta)| cannot be negative. But you can write it as \pm cos(\theta)| because cos(\theta) itself can be positive or negative.

IF \theta= (5/6)\pi then cos(\theta)= cos((5/6)\pi)= -\sqrt{3}/2 so |cos(\theta)|= - cos(\theta)= \sqrt{3}/2.

 

[songoku]

Hi njama and Mr. HallsofIvy

This is my interpretation after reading your posts.

\int_{0}^{0.5}\frac{xdx}{\sqrt{1-x^{2}}} can turn out to be:

1.\int_{0}^{\frac{\pi}{6}}\sin \theta\;d\theta\;\text{or}

2.\int_{\pi}^{\frac{5}{6}\pi}-\sin \theta\;d\theta

Am I right? Or maybe there are any other forms with different upper and lower bounds?

Thanks