Sevaji , your help is needed

by Rishi Sun Apr 14, 2013 9:43 pm

Evaluate

1
∫sqrt(x)/(x+1)
0

Here is my solution. But I got stuck. By intuition, I figured out the right answer on the multiple choice quiz. Please tell me where I went wrong.

Let x = u^2

By implicit differentiation dx = 2udu

∫ (sqrt(u^2)/(u^2 + 1) ) * (2udu)

2∫((u^2)*du)/((U^2) + 1)

2 ∫ (1*du)/(1 + (1/(u^2))

=2 (arctan(1/(u)))

Since x = u^2 then 1/u = 1/sqrt(x)

the original problem reduces to

(2 *arctan(1/sqrt(1)) - (2 * arctan(1/sqrt(0)))

I know this is utterly wrong.

Where did I go wrong?

by MaxEntropy_Man Sun Apr 14, 2013 10:38 pm

let me give you my solution and then i'll look at yours. i got 2- pi/2. i suspect there are many ways to skin this cat, but here is mine.

i substituted x = (tan y)^2. if you do that the integral reduces to integral [2 (tan y)^2, y] which can be written as integral[2((sec y)^2-1), y]. this is easy to integrate and is just 2 (tan y - y). you can either change the limits to y at this point, or rewrite the answer in terms of the original x in which case the answer is 2 (sqrt(x) - arctan(sqrt(x)). you can now apply the limits and get 2- pi/2.

p.s: how did you post symbolic math?

by Rishi Sun Apr 14, 2013 10:45 pm

MaxEntropy_Man wrote:let me give you my solution and then i'll look at yours. i got 2- pi/2. i suspect there are many ways to skin this cat, but here is mine.

i substituted x = (tan y)^2. if you do that the integral reduces to integral [2 (tan y)^2, y] which can be written as integral[2((sec y)^2-1), y]. this is easy to integrate and is just 2 (tan y - y). you can either change the limits to y at this point, or rewrite the answer in terms of the original x in which case the answer is 2 (sqrt(x) - arctan(x)). you can now apply the limits and get 2- pi/2.

p.s: how did you post symbolic math?

I just copied and pasted the integral sign from the Coursera.org website.

by MaxEntropy_Man Sun Apr 14, 2013 10:52 pm

you were ok till this step:

2∫((u^2)*du)/((u^2) + 1). after that you could split this up using partial fractions as,

2[(1+u^2)-1]/(1+u^2). splitting that up you get,

2-2/(1+u^2). integrating that you get, 2(u-arctan(u)). inverting your original substitution you get, 2(sqrt(x)-arctan(sqrt(x))) which is the same result i got. you can now apply the limits.

p.s: in my original post there was a typo which i fixed subsequently. the argument of the arctan in the final answer is sqrt(x), not x as i had originally posted.

by MaxEntropy_Man Sun Apr 14, 2013 11:02 pm

i also found your error. your answer for the integral in terms of u, i.e.
arctan (1/u) is wrong. you can see that for yourself by differentiating your answer. you get -1/(1+u^2), and not your integrand.

by Rishi Mon Apr 15, 2013 12:13 am

MaxEntropy_Man wrote:let me give you my solution and then i'll look at yours. i got 2- pi/2. i suspect there are many ways to skin this cat, but here is mine.

i substituted x = (tan y)^2. if you do that the integral reduces to integral [2 (tan y)^2, y] which can be written as integral[2((sec y)^2-1), y]. this is easy to integrate and is just 2 (tan y - y). you can either change the limits to y at this point, or rewrite the answer in terms of the original x in which case the answer is 2 (sqrt(x) - arctan(sqrt(x)). you can now apply the limits and get 2- pi/2.

p.s: how did you post symbolic math?

I do not understand your notation integral [ 2 (tany) ^2, y ]

I substituted x = tan^2 U in my method.

It worked like a charm.

Thanks.

I am just curious.

What made you think of using tan^2 in the substitution?

Is it because the original problem had 1 + x in the denominator and that reminded you of using the equality i + tan^2 (u) = sec^2 (u) ?

by MaxEntropy_Man Mon Apr 15, 2013 6:51 am

Rishi wrote:

I do not understand your notation integral [ 2 (tany) ^2, y ]

by that i meant the integral of 2 (tan y)^2 with respect to y.

Rishi wrote:I substituted x = tan^2 U in my method.

It worked like a charm.

Thanks.

I am just curious.

What made you think of using tan^2 in the substitution?

Is it because the original problem had 1 + x in the denominator and that reminded you of using the equality i + tan^2 (u) = sec^2 (u) ?

yes. the fact that i could get a sec^2 downstairs which had the potential of canceling with the sec^2 upstairs resulting from differentiation was intriguing. so i tried it. your own method is just fine. had you recognized that u^2/(u^2+1) can be written as (u^2+1-1)/(u^2+1), and therefore 1-1/(u^2+1), you would have been further along.

by Seva Lamberdar Mon Apr 15, 2013 8:48 am

Rishi wrote:
Evaluate

1
∫sqrt(x)/(x+1)
0

Here is my solution. But I got stuck. By intuition, I figured out the right answer on the multiple choice quiz. Please tell me where I went wrong.

Let x = u^2

By implicit differentiation dx = 2udu

∫ (sqrt(u^2)/(u^2 + 1) ) * (2udu)

2∫((u^2)*du)/((U^2) + 1)

2 ∫ (1*du)/(1 + (1/(u^2))

=2 (arctan(1/(u)))

Since x = u^2 then 1/u = 1/sqrt(x)

the original problem reduces to

(2 *arctan(1/sqrt(1)) - (2 * arctan(1/sqrt(0)))

I know this is utterly wrong.

Where did I go wrong?

Rishi, the best thing to do in such problems is to substitute a suitable variable for the original variable, like you already did, so that the problem becomes simpler and easier to solve. It's more like a trial and error method, i.e. if you don't get the answer first time (using first alternate variable), try another one and then try another one......

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