# Big trigonometric substitution problem.

• Feb 19th 2011, 05:35 PM
Wolvenmoon
Big trigonometric substitution problem.
Okay, it's the last problem I have to do in the section and it stumped me.

It's the integral of ((4-x)/x)^(1/2)dx

The book says to let x=u^2.

The answer is 4*arcsin(x^(1/2)/2) + x^(1/2) * (4-x)^(1/2) + C.

No matter what I try I am not getting close to this answer. I know I'm supposed to get it into two pieces, one with form integral du/(a^2-u^2) .

I start by letting u = x^(1/2). u^2=x. du= 1/2*x^(-1/2)

So this gives me 2*integral square root of (4-u^2)/u^2 du.

I then have to substitute. Since the integral is in the form of the square root of a^2-x^2 I use u=a*sin(t) or U=2*sin(t) and du=2*cos(t).

This gives me the integral of 2*((4-4sin^2(t))/(4sin^2(t))^(1/2) * 2*cos(t).

I think I have it right down to this point. Then when I retry the problem I keep trying different things from here on out.

Sorry for the TI-89 format. I'm going to try to learn LaTex in the next week or so.

Thanks for your help!
• Feb 19th 2011, 07:23 PM
tonio
Quote:

Originally Posted by Wolvenmoon
Okay, it's the last problem I have to do in the section and it stumped me.

It's the integral of ((4-x)/x)^(1/2)dx

The book says to let x=u^2.

The answer is 4*arcsin(x^(1/2)/2) + x^(1/2) * (4-x)^(1/2) + C.

No matter what I try I am not getting close to this answer. I know I'm supposed to get it into two pieces, one with form integral du/(a^2-u^2) .

I start by letting u = x^(1/2). u^2=x. du= 1/2*x^(-1/2)

So this gives me 2*integral square root of (4-u^2)/u^2 du.

I then have to substitute. Since the integral is in the form of the square root of a^2-x^2 I use u=a*sin(t) or U=2*sin(t) and du=2*cos(t).

This gives me the integral of 2*((4-4sin^2(t))/(4sin^2(t))^(1/2) * 2*cos(t).

I think I have it right down to this point. Then when I retry the problem I keep trying different things from here on out.

Sorry for the TI-89 format. I'm going to try to learn LaTex in the next week or so.

Thanks for your help!

$\displaystyle{\int\sqrt{\frac{4-x}{x}}\,dx\,,\,\,u^2=x\Longrightarrow 2u\,du=dx}$ , and

$\displaystyle{\int\sqrt{\frac{4-u^2}{u^2}}\,2u\,du=2\int\sqrt{4-u^2}\,du=4\int\sqrt{1-\left(\frac{u}{2}\right)^2}}\,du}$

Put now $\displaystyle{\frac{u}{2}=\sin t\Longrightarrow du=2\cos t\,dt\Longrightarrow 4\int\sqrt{1-\left(\frac{u}{2}\right)^2}}\,du=8\int \cos^2t\,dt=}$

$\displaystyle{=4(t+sin t\cos t)+C=4\left(\arcsin \frac{u}{2}+\frac{u}{2}\sqrt{1-\left(\frac{u}{2}\right)^2\right)+C$ and etc.

Tonio