Trouble with Integral of x*arcsec(squareroot(x)) dx

[kilroymcb]

Hello
This is my first post here and I hope you'll bare wtih me on this.
The problem is:
Int X*arcsec(sqrt(x))dx.
The only approach that seemed to make sense on this problem was to use Integration by parts to get the problem into some form of integral form. As I do not know the integral of arcsec, I chose my dV to be x, and my U to be arcsec(sqrt(x)).
Then, I used the regular formula for integration by parts (UV-int(v)(dU)). Of course, V was x^2/2 and du was 1/ sqrt(2)*sqrt((sqrt(x))^2 - 1.

The problem at this point is that over in the new integral, I've got what would ordinarily be an easy Trig substitution, except that I have that sqrt (x) for my variable term! If I let sqrt(x) = sec (#) then I get x by itself by taking ^2 of sec, then for dx I get something I dont know... Also, I see this Integration by parts as getting worse and worse... not better.

I'd appreciate any assistance on this.[/code]

 

[galactus]

You have: \(\displaystyle \L\\\int{xsec^{-1}(x)}dx\)

Now, use parts: Let \(\displaystyle u=sec^{-1}(x), \;\ dv=x, \;\ du=\frac{1}{x\sqrt{x^{2}-1}}, \;\ v=\frac{1}{2}x^{2}\)

Give those a try and you should end up with the correct evaluation, which is:

\(\displaystyle \L\\\frac{1}{2}x^{2}sec^{-1}(x)-\frac{1}{2}\sqrt{x^{2}-1}\)


See if you can get there.

EDIT: I am sorry, I just noticed the square root in your sec. Give 'er a go anyway.

 

[kilroymcb]

You have: \(\displaystyle \L\\\int{xsec^{-1}(x)}dx\)

Now, use parts: Let \(\displaystyle u=sec^{-1}(x), \;\ dv=x, \;\ du=\frac{1}{x\sqrt{x^{2}-1}}, \;\ v=\frac{1}{2}x^{2}\)

Give those a try and you should end up with the correct evaluation, which is:

\(\displaystyle \L\\\frac{1}{2}x^{2}sec^{-1}(x)-\frac{1}{2}\sqrt{x^{2}-1}\)


See if you can get there.

EDIT: I am sorry, I just noticed the square root in your sec. Give 'er a go anyway.

Hello and thanks for the prompt response. However, if you'll look over my original post, you'll see that I tried exactly that and I'm stuck. Any other suggestions?

 

[skeeter]

let's see if this may work ...

let \(\displaystyle \L y = arcsec\sqrt{x}\)

\(\displaystyle \L \sec{y} = \sqrt{x}\)

\(\displaystyle \L x = \sec^2{y}\)

\(\displaystyle \L dx = 2\sec^2{y}\tan{y} dy\)

so ...

\(\displaystyle \L \int x \cdot arcsec\sqrt{x} dx\)

\(\displaystyle \L \int \sec^2{y} \cdot y \cdot 2\sec^2{y}\tan{y} dy\)

\(\displaystyle \L 2 \int y \cdot \sec^3{y} \sec{y} \tan{y} dy\)

let \(\displaystyle \L u = y\) ... \(\displaystyle \L du = dy\)

\(\displaystyle \L dv = \sec^3{y} \sec{y} \tan{y} dy\)

\(\displaystyle \L v = \frac{\sec^4{y}}{4}\)

\(\displaystyle \L y \cdot \frac{\sec^4{y}}{2} - \int \frac{\sec^4{y}}{2} dy\)

\(\displaystyle \L y \cdot \frac{\sec^4{y}}{2} - \frac{1}{2} \int \sec^2{y}(\tan^2{y}+1)dy\)

\(\displaystyle \L y \cdot \frac{\sec^4{y}}{2} - \frac{1}{2} \int \sec^2{y}\tan^2{y}dy - \frac{1}{2} \int \sec^2{y} dy\)

\(\displaystyle \L y \cdot \frac{\sec^4{y}}{2} - \frac{\tan^3{y}}{6} - \frac{\tan{y}}{2} + C\)

now ... since \(\displaystyle \L x = sec^2{y}\),
\(\displaystyle \L x = 1 + \tan^2{y}\),
and
\(\displaystyle \L \tan{y} = (x - 1)^{\frac{1}{2}\)

so ... the final antiderivative is

\(\displaystyle \L \frac{x^2}{2} arcsec\sqrt{x} - \frac{(x-1)^{\frac{3}{2}}}{6} - \frac{(x-1)^{\frac{1}{2}}}{2} + C\)

of course, the only way to check is to take the derivative of my solution, simplify, and see if we end up with the original integrand.

 

[kilroymcb]

Wow! That looks reasonable... let me run through that myself and see if I get anything wonky.

thanks!

 

[stapel]

I hope you'll bare wtih me on this....

You mean "bear with me", right...? Cuz I ain't gettin' nekkid.... :wink:

Eliz.

 

[kilroymcb]

I hope you'll bare wtih me on this....

You mean "bear with me", right...? Cuz I ain't gettin' nekkid.... :wink:

Eliz.

You guys rock! Not only do I get calculus help, but english lessons as well!