How can I solve the integral of Sqr(1-4x^2) using trig substitution?

[lemurs]

Ok guys here we go. this isI just a suggested question from our teacher in prep for our final. it a integral one. trig substion the prob I am having is near the end.. all the substition works great was down to the last bit and for some reason I must be doing the back substition wrong. so here it is.

orginal question

Integral of Sqr(1-4x^2) dx

I do the substion of x=1/2 sin @
(@= theta)
and do all that stuff.

after I integrat i have 1/4 @ + 1/8 sin2@

so i do the back sub for just @ i get arcsine 2x which is correct but some reason i don;t get the back end right...

some how 1/8 sin2@ = 1/2*x*sqr(1-4x^2)

so if you could fight out the why i am going wrong and screwing up could you tell me.

 

[dextercioby]

You've made a mistake in computing the integral of $\cos^{2}\theta$

Daniel.

 

[lemurs]

whawt was I supposed do then.. as far as i know i should use the fact that cos @^2 =1/2(1+cos 2@)
if i did that wrong then what is the correct way?

 

[HallsofIvy]

$$\int \sqrt{1- 4x^2}dx$$
Let $sin(\theta)= 2x$. Then $cos(\theta)d\theta= 2 dx$ so that $\frac{1}{2}cos(\theta)d\theta= xdx$ and $\sqrt{1- 4x^2}= \sqrt{1- sin^2(\theta)}= cos(\theta)d\theta$.

The integral becomes
$$\int cos^2(\theta)d\theta$$
Yes, you are correct that $cos^2(\theta)= \frac{1}{2}(1+cos(2\theta)$ so that integral is
$$\frac{1}{2}(1+ cos(2\theta)d\theta$$
[tex]= \frac{1}{2}(\theta+ \frac{1}{2} sin(2\theta)+ C[/itex]
Since $sin(\theta)= 2x$,$x= arcsin(2x)$ and
$sin(2\theta)= 2sin(\theta)cos(\theta)= 2x\sqrt{1-4x^2}$
($cos(\theta)= \sqrt{1- sin^2(\theta)}= \sqrt{1- 4x^2}$)
this is
[tex]\frac{1}{2}(2arcsin(2x)+ x\sqrt{1-4x^2}+ C[/itex]

 

[lemurs]

ok just want to m ake sure I get this. to solve the sin (2@) you used an Identity that sin 2@ = 2sin@cos@

and you can find cos cause we know what sin is.
correct?

 

[dextercioby]

Yes, correct.

Daniel.