Posted by Dartchen on August 29, 2002 at 17:46:15:
In Reply to: Re: Need help desperately - seems like a simple integral? posted by Ryan on August 29, 2002 at 15:27:04:
Hmmm...is it possible to use the reverse of the product rule for integration in this case? Just got done reading an article http://mathcentral.uregina.ca/QQ/database/QQ.09.99/shah3.html which has a similar problem that resulted in an answer with a Polylog. Now I'm intrigued. It seems that math programs are coming up with an answer but I'm curious how they are arriving at that answer. Any ideas?
: I used the product rule--> D of U * V = U*V' + V*U'
: so now
: 2* integral [ u tan u du ] -->
: 2* (1/2U^2 * tanu + u * ln | sec u | ) +C
: : Good deal...I like seeing how other people would work the problem. I agree with the correction that gets us to 2* Integral u*tan(u)du. In your original solution based off of u^2 vs u there was a jump on the last step to a final solution. Which formula/rule did you use to get there and would it still apply if we use u vs u^2?
: : Thanks!
: : : CORRECTION
: : : 2u = dx
: : : 2* integral [ u tanu du]
: : : : ill give your prob. a try--> I have been teaching myself calculus so I am not sure if my answer is right....use your own judgement!
: : : : integral ( tan(x^1/2))
: : : : u= x^1/2
: : : : du = (1/2)x^(-1/2) dx
: : : : 2du = x^-1/2 dx
: : : : 2u^2 du = dx
: : : : 2* integral [( u^2)*(tan u ) du ]
: : : : 2 * (1/3)u^3 tan u + u^2 ln |sec u | + C
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