## anonymous 5 years ago Use an appropriate change of variable to evaluate the convergent improper integral 1/[(x^0.5)(1+x)] what should i use?

1. anonymous

the answer is (pi) but i dunno how should i get to the answer

2. anonymous

Is it dy/dx = 1/[(x^0.5)(1+x)] ? Move the dx to the right side, integrate the left and you get "y", integrate the right side now and try using a trig substitution...

3. anonymous

sorry..my bad... 1st time use this website...but i found this fun ^ ^ the question is based on improper integral. $\int\limits\limits_{0}^{\infty} 1\div [ \sqrt{x} (1+x) ]$

4. anonymous

Okay, for the right hand side, you're going to use a u-substitution (try u = sqrt(x)) and after the sub, you should see a pretty clean inverse trig function that pops out.

5. anonymous

Then, finding evaluating at the limits will be an easy task. :)

6. anonymous

*eliminate "finding" :P typo.

7. anonymous

thanks for helping but i cant get through with the trig part.. i cant see which i use to substitute i stuck at $1\div (u + u ^{3})$ :(

8. anonymous

I've been scratching my head at this one for a while, but I just got it. :) You've got du=1/(2sqrt(x)) dx, right? You've gotta multiply the left by 2sqrt(x) to get dx = 2√x du! Plug THAT into your integral and you'll get 2 * ∫ 1/(1+u^2) du. That should be much easier to solve!

9. anonymous

hm...i getting 0 at the end .. hm... odd.. the answer i getting from my lecture is Pi what is the answer u getting? i get in the end $\tan^{-1} u$

10. anonymous

Exactly, you're just forgetting to multiply by 2. So, your final expression in the end is 2 * arctan√x from 0 to ∞. The value of arctan(√x) as x approaches infinity is π/2, and as x approaches 0, is 0! So you get 2* (π/2-0) after applying the limits and the final value is π.

11. anonymous

damn i keep doing so stupid mistake...i might need my brain rest for a while..XD..it's been 5 hours for me ... thanks..u r awesome.. ^ ^