Help with integral

**angel.white** · March 3rd 2008, 12:40 AM

Can anyone do this integral for me? I am not convinced my answer is correct :/

$\int \sqrt{1+2e^{2t}}~dt$

Here is my work:

---------------------------------------------
SUBSTITUTE:
$\frac{1}{\sqrt{2}}~tan(a)=e^t$

$\frac{1}{\sqrt{2}}~sec^2(a)~da=e^t~dt$

$\frac{1}{\sqrt{2}e^t}~sec^2(a)~da=dt$

$cot(a)~sec^2(a)~da=dt$

$csc(a)~sec(a)~da=dt$
---------------------------------------------

= $\int csc(a)~sec(a) \sqrt{1+2\left(\frac{tan(a)}{\sqrt{2}}\right)^2}~d a$

= $\int csc(a)~sec(a) \sqrt{1+tan^2(a)}~da$

= $\int csc(a)~sec^2(a)~da$

= $\int csc(a)~(1+tan^2(a))~da$

= $\int csc(a)~da + \int \frac{sin(a)}{cos^2(a)}~da$

= $-ln| csc(a)+cot(a) | + sec(a)~da$

= $-ln\left|\frac{1+\sqrt{2e^{2t}+1}}{\sqrt{2}e^t}\rig ht|+(2e^{2t}+1)^{-1/2} + C$

I've done it twice and gotten this answer, but it just seems kind of ugly and unlikely, and my calculator told me a different answer

**mr fantastic** · March 3rd 2008, 02:34 AM

Originally Posted by angel.white

Can anyone do this integral for me? I am not convinced my answer is correct :/

$\int \sqrt{1+2e^{2t}}~dt$

Here is my work:

---------------------------------------------
SUBSTITUTE:
$\frac{1}{\sqrt{2}}~tan(a)=e^t$

$\frac{1}{\sqrt{2}}~sec^2(a)~da=e^t~dt$

$\frac{1}{\sqrt{2}e^t}~sec^2(a)~da=dt$

$cot(a)~sec^2(a)~da=dt$

$csc(a)~sec(a)~da=dt$
---------------------------------------------

= $\int csc(a)~sec(a) \sqrt{1+2\left(\frac{tan(a)}{\sqrt{2}}\right)^2}~d a$

= $\int csc(a)~sec(a) \sqrt{1+tan^2(a)}~da$

= $\int csc(a)~sec^2(a)~da$

= $\int csc(a)~(1+tan^2(a))~da$

= $\int csc(a)~da + \int \frac{sin(a)}{cos^2(a)}~da$

= $-ln| csc(a)+cot(a) | + sec(a)~da$

= $-ln\left|\frac{1+\sqrt{2e^{2t}+1}}{\sqrt{2}e^t}\rig ht|+(2e^{2t}+1)^{-1/2} + C$

I've done it twice and gotten this answer, but it just seems kind of ugly and unlikely, and my calculator told me a different answer

Note that:

1. $\sec a = \sqrt{2 e^{2t} + 1}$ , NOT $(2e^{2t}+1)^{-1/2} = \frac{1}{\sqrt{2 e^{2t} + 1}}$ :

$\tan a = \sqrt{2} e^t \Rightarrow \tan^2 a = 2 e^{2t} \Rightarrow 1 + \tan^2 a = 1 + 2 e^{2t} \Rightarrow \sec^2 a = 2e^{2t} + 1$ .

2. $\ln\left|\frac{1+\sqrt{2e^{2t}+1}}{\sqrt{2}e^t}\ri ght| = \ln \left( 1+\sqrt{2e^{2t}+1} \right) - \ln(\sqrt{2} e^t)$

$= \ln \left( 1+\sqrt{2e^{2t}+1} \right) - \ln(e^t) - \ln \sqrt{2} = \ln \left( 1+\sqrt{2e^{2t}+1} \right) - t - \ln \sqrt{2}$ .

The $\ln \sqrt{2}$ can be soaked up into C. Then you get an answer that will almost agree with what your calculator is probably telling you ...... The snag is that there shouldn't be a negative in front of you log .....

In other words, the answer is

$\ln \left( 1+\sqrt{2e^{2t}+1} \right) - t + \sqrt{2 e^{2t} + 1} + C$

NOT $-\ln \left( 1+\sqrt{2e^{2t}+1} \right) + t + \sqrt{2 e^{2t} + 1} + C$ .

I can't see off-hand where your unwanted negative has come from ......

**angel.white** · March 3rd 2008, 03:08 AM

Originally Posted by mr fantastic

Note that:

1. $\sec a = \sqrt{2 e^{2t} + 1}$ , NOT $(2e^{2t}+1)^{-1/2} = \frac{1}{\sqrt{2 e^{2t} + 1}}$ :

$\tan a = \sqrt{2} e^t \Rightarrow \tan^2 a = 2 e^{2t} \Rightarrow 1 + \tan^2 a = 1 + 2 e^{2t} \Rightarrow \sec^2 a = 2e^{2t} + 1$ .

2. $\ln\left|\frac{1+\sqrt{2e^{2t}+1}}{\sqrt{2}e^t}\ri ght| = \ln \left( 1+\sqrt{2e^{2t}+1} \right) - \ln(\sqrt{2} e^t)$

$= \ln \left( 1+\sqrt{2e^{2t}+1} \right) - \ln(e^t) - \ln \sqrt{2} = \ln \left( 1+\sqrt{2e^{2t}+1} \right) - t - \ln \sqrt{2}$ .

The $\sqrt{2}$ can be soaked up into C. Then you get an answer that will almost agree with what your calculator is probably telling you ...... The snag is that there shouldn't be a negative in front of you log .....

In other words, the answer is

$\ln \left( 1+\sqrt{2e^{2t}+1} \right) - t + \sqrt{2 e^{2t} + 1} + C$

NOT $-\ln \left( 1+\sqrt{2e^{2t}+1} \right) + t + \sqrt{2 e^{2t} + 1} + C$ .

I can't see off-hand where your unwanted negative has come from ......

Thank you, it's been a very long couple of weeks >.<

the secant thing, I just made a careless error on, upon review, I see your answer is correct.

And good eye on pulling out the t in the denominator, that's definitely why my answer was so far off from theirs. You'd think I'd remember to check such things :/

The negative on the derivative of csc(a) came because I don't have that function's integral memorized, so I just went to look it up by typing it into the calculator. It told me:

$\int csc(x) = -\ln | cot(x) + csc(x) | + C$

out of annoyance with myself for not just figuring it out, I went to figure it out just now... and I'm having a tough time of it, can you show me how to it?

**mr fantastic** · March 3rd 2008, 03:25 AM

Originally Posted by angel.white

Thank you, it's been a very long couple of weeks >.<

the secant thing, I just made a careless error on, upon review, I see your answer is correct.

And good eye on pulling out the t in the denominator, that's definitely why my answer was so far off from theirs. You'd think I'd remember to check such things :/

The negative on the derivative of csc(a) came because I don't have that function's integral memorized, so I just went to look it up by typing it into the calculator. It told me:

$\int csc(x) = -\ln | cot(x) + csc(x) | + C$

out of annoyance with myself for not just figuring it out, I went to figure it out just now... and I'm having a tough time of it, can you show me how to it?

Go to the relevant link here for the easy way. The tougher way uses the Weiestrass substitution.

I know why you have the negative, the thing is that it's not wanted in the correct answer ...... Can't see off-hand why it's there when it shouldn't be ......

**Krizalid** · March 3rd 2008, 05:08 AM

Why make this thing so hard?

Originally Posted by angel.white

Can anyone do this integral for me? I am not convinced my answer is correct :/

$\int \sqrt{1+2e^{2t}}~dt$

Substitute $z^2=1+2e^{2t},$ the integral becomes $\int {\frac{{z^2 }} {{z^2 - 1}}\,dz} = z + \frac{1} {2}\ln \left| {\frac{{z - 1}} {{z + 1}}} \right| + k.$

Just back substitute.

**mr fantastic** · March 3rd 2008, 12:01 PM

Originally Posted by Krizalid

Why make this thing so hard?

Substitute $z^2=1+2e^{2t},$ the integral becomes $\int {\frac{{z^2 }} {{z^2 - 1}}\,dz} = z + \frac{1} {2}\ln \left| {\frac{{z - 1}} {{z + 1}}} \right| + k.$

Just back substitute.

Yes, I'd thought of that. And of course an answer in this form immediately suggests a simplification using the definition of the inverse tanh function ....

But hard or easy, all valid methods should lead to equivalent answers.

I generally like to find the mistake in the method used by the person asking the question (unless the method follows the Highway to Hell) before suggesting other approaches. I find that gives the person greater ownership of the solution .....

Thread: Help with integral

LinkBack

Thread Tools

Display

Help with integral

Similar Math Help Forum Discussions

Use Cauchy's integral theorem for derivatives to evaluate the contour integral.

Graph the region of integration and evaluate the double integral (integral from 0 to

Explaining result: comparing line integral and two-form integral

write [integral of] dx/SQRT(sinx) in terms of an elliptic integral

Evaluating a double integral to find its signle integral.

Search Tags