# An effective substitution

1. Dec 26, 2004

### brendan_foo

Hi guys, could someone just suggest a variable substitution for me...just to get the ball rolling :

$$\int_{-1}^{1} \sqrt{1-x^2} - x^2 (\sqrt{1-x^2}) - (1-x^2)^\frac{3}{2} dx$$

Whenever i've seen $\sqrt{1-x^2}$ (and its usually the reciprocal of), I've used trig substitutions to wind up with an inverse trigonometrical function as the output, and then evaluating the integral. Either way, would something like $x = \sin{\theta}$ be a good choice here (or maybe $x = \csc{\theta}$), or can it just be done in a regular fashion? Any hints would be great, nothing revealing but a nudge in the right direction would be ace, if possible.

cheers guys

2. Dec 26, 2004

### hedlund

Use this $$\int_{-1}^{1} \sqrt{1-x^2} - x^2 (\sqrt{1-x^2}) - (1-x^2)^\frac{3}{2} \ dx = \int_{-1}^{1} \sqrt{1-x^2} \ dx - \int_{-1}^{1} x^2\sqrt{1-x^2} \ dx - \int_{-1}^{1} \left( 1- x^2\right)^{\frac{3}{2}} \ dx$$

And make substituion for each integral ...

3. Dec 26, 2004

### Benny

For this one just do some factoring.

$$\int\limits_{ - 1}^1 {\sqrt {1 - x^2 } } - x^2 \left( {\sqrt {1 - x^2 } } \right) - \left( {1 - x^2 } \right)^{\frac{3}{2}} dx$$

$$= \int\limits_{ - 1}^1 {\sqrt {1 - x^2 } \left( {1 - x^2 - \left( {1 - x^2 } \right)} \right)} dx$$

$$= \int\limits_{ - 1}^1 {\left( { - 2x^2 } \right)\sqrt {1 - x^2 } } dx$$

I am not too sure about the substitution though. Maybe try some standard trig substitutions and then integrate by parts if needed?

Edit: I just tried using trig substitution with what I came up with. It ends up requiring quite a bit of working so it may not be the best method to use.

Last edited: Dec 26, 2004
4. Dec 26, 2004

### Muzza

You know, (1 - x^2 - (1 - x^2)) = (1 - x^2 - 1 + x^2) = 0.

5. Dec 26, 2004

### dextercioby

This integral
$$\int_{-1}^{+1} [\sqrt{1-x^{2}} - x^{2} (\sqrt{1-x^{2}}) - (1-x^{2})^{\frac{3}{2}}] dx$$

is dealt nicely by the substitution $x\rightarrow \sin u$,under which the limits of integration transform as follows:
$$-1\rightarrow -\frac{\pi}{2} ; +1\rightarrow +\frac{\pi}{2}$$
,and your integral becomes:
$$\int_{-\frac{\pi}{2}}^{+\frac{\pi}{2}} (\cos u -\sin^{2}u\cos u-\cos^{3}u)\cos u du$$
,which can be evaluated quite easily.
Split it in three integrals.Denote your initial integral by 'I'.Then:
$$I=I_{1}-I_{2}-I_{3}$$
,where
$$I_{1}=\int_{-\frac{\pi}{2}}^{+\frac{\pi}{2}} \cos^{2}u du$$
$$I_{2}=\int_{-\frac{\pi}{2}}^{+\frac{\pi}{2}} \sin^{2}u\cos^{2}u du$$
$$I_{3}=\int_{-\frac{\pi}{2}}^{+\frac{\pi}{2}} \cos^{4}u du$$.

Use the fact that $\cos^{4}u=\cos^{2}u-\cos^{2}u\sin^{2}u$
to express
$$I_{3}=I_{1}-I_{2}$$
,from where u find that your initial integral (I) is:
$$I=0$$

Daniel.

6. Dec 26, 2004

### brendan_foo

Hi, I managed to come to the integral(s) that you arrived to in the previous post, but I think I sucked up those integrations.

I know the final integral should be evaluated to 1/5, but its a really long winded little thing. I used $x = \sin(\theta)$ but the calculations went on for ages.

Could you use reciprocal trig functions, like cosec(x) or something and use those trig identies? or does that just make it un-necessary complication?

Cheers guys, much appreciated

tell me if you can arrive at one-fifth.

7. Dec 26, 2004

### dextercioby

I just showed you that,in order to evluate the initial integral,you have to make an obviuos substitution and no integration,as the result is proven to be zero by a simple identity involving trig.functions.

Consider the integrals:
$$I_{s}=:\int_{-\frac{\pi}{2}}^{+\frac{\pi}{2}} \sin^{2}x dx$$
$$I_{c}=:\int_{-\frac{\pi}{2}}^{+\frac{\pi}{2}} \cos^{2}x dx$$

1.Use the fundamental formula of circular trigonometry to find
$$I_{c}+I_{s}=\int_{-\frac{\pi}{2}}^{+\frac{\pi}{2}} 1 dx =x|_{-\frac{\pi}{2}}^{+\frac{\pi}{2}}=\pi$$
2.Use the identity:
$$\cos^{2} x-\sin^{2} x=\cos 2x$$
,to find that:
$$I_{c}-I_{s}=\int_{-\frac{\pi}{2}}^{+\frac{\pi}{2}} \cos 2x dx$$
.Under the obvious substitution $2x\rightarrow u$,under which the limits transform as follows:
$$-\frac{\pi}{2}\rightarrow -\pi ;+\frac{\pi}{2}\rightarrow +\pi$$
,u're gettin'
$$I_{c}-I_{s}=\frac{1}{2}\int_{-\pi}^{+\pi} \cos u du =\frac{1}{2}\sin u|_{-\pi}^{+\pi}=0$$
So:
$$I_{c}=I_{s}=\frac{\pi}{2}$$
.And should be the first integral $I_{1}$
For the second,use the trigonometrical identity
$$\sin^{2}x\cos^{2}x =(\frac{1}{2}\sin 2x)^{2}=\frac{1}{4}\sin^{2}2x$$.
And then the integral,by a substitution $2x\rightarrow u$ could be put in relation with $I_{s}$ computed above.

Daniel.

Last edited: Dec 26, 2004
8. Dec 26, 2004

### Muzza

You don't have to involve any trigonometric functions, see Benny's post and my correction to it.

9. Dec 26, 2004

### dextercioby

You're right,but the trig.stuff is needed to evaluate each integral in term.He was convinced that the final result would be zero,but he asked whether the last integral (I_{3} as i depicted it) was 1/5 or not...

Daniel.

10. Dec 26, 2004

### brendan_foo

Hey are any of you guys on an IM facility, like MSN or yahoo...wouldn't mind having a proper chat with someone, I can see that in comparison to many, i am waaaaay behind.

brendan_online@hotmail.com if you could talk to me on MSN

thanks for your patience
Cheers guys

Last edited: Dec 26, 2004
11. Dec 26, 2004

### Muzza

Aha, I interpreted "final integral" as "I".

12. Dec 26, 2004

### Benny

Oops, I made an error with my simplification.