How Do You Integrate the Square Root of Tangent?

[Jameson]

Yes, it looks like it could be equilvalent.

 

[GCT]

If you're doubting the calculations exposed above and,incidentally,not shortcut,why won't you put a pencil in your precious hand and grab a piece of paper and do it...?

Sides,Maple is a known f***-up.

Actually I was working on this problem myself and thought it could be done by partial fractions after arriving at

$$2\int \frac{t^{2}}{1+t^{4}}\dt$$, I wasn't aware of the trick to progress towards the next step though

Your final form seems fine, however, I'll need to do some research on what makes such a partial fraction capable of integration (just finished int. calc this spring). Perhaps when I find the time, I'll try solving it p&p, just thought it could be done through Maple or Matlab. Isn't this a table integral (the OP's integral)?

It would also be appropriate for the final answer to be posted conclusively, if your final form proves to be right, good for you...and all of us, if not (I would be surprised if it isn't), than let's move on to see exactly why.

 

[dextercioby]

Yes.It's [1]:

$$\int \frac{x^{2}}{A+Bx^{4}} \ dx=\left\{\begin{array}{c} \frac{1}{4B\alpha\sqrt{2}}\left(\ln\frac{x^{2}-\alpha x\sqrt{2}+\alpha^{2}}{x^{2}+\alpha x\sqrt{2}+\alpha^{2}}+2\arctan\frac{\alpha x\sqrt{2}}{\alpha^{2}-x^{2}}\right) \ \mbox{if} \ AB>0\\ -\frac{1}{4B\alpha'}\left(\ln\frac{x+\alpha'}{x-\alpha'}-2\arctan\frac{x}{\alpha'}\right) \ \mbox{if} \ AB<0 \end{array}\right$$

,where

$$\alpha=\sqrt[4]{\frac{A}{B}}$$

$$\alpha'=\sqrt[4]{\frac{-A}{B}}$$.

Daniel.

-------------------------------------------------
[1]G & R,5-th edition,CD version,Academic Press,1996,formula 2.132,3)

 

[GCT]

alright, I'll try the good o'l fashioned way. Somebody please check my work,

$$I=2 \int \frac{t^{2}~dt}{(1+ \sqrt{2}t+t^{2})(1- \sqrt{2}t +t^{2})}$$

Aside

$$\frac{t^{2}}{(1+ \sqrt{2}t+t^{2})(1- \sqrt{2}t +t^{2})}$$=

$$\frac{Ax+B}{1+ \sqrt{2}t +t^{2}}~+~ \frac{Cx+D}{1- \sqrt{2}t+t^{2}}$$

$$t^{2}=(Ax+B)(1- \sqrt{2}t +t^{2})+(Cx+D)(1+ \sqrt{2}t +t^{2})$$

$$Ax- \sqrt {2} Axt +Axt^{2} +B - \sqrt{2}Bt +Bt^{2} +Cx + \sqrt{2}Cxt +Cxt^{2} +D + \sqrt{2}Dt +Dt^{2}$$

=

$$(Ax+B+Cx+D)t^{2} + (- \sqrt{2} AX - sqrt{2}B + \sqrt{2} Cx + \sqrt{2} D)t +Ax+B+Cx+D$$



1)$$Ax+B+Cx+D=1$$
2)$$- \sqrt{2} Ax - \sqrt{2} B + \sqrt{2} Cx + \sqrt{2}D=0$$
3)$$Ax+B+Cx+D=0$$

It seems that we have a problem here between 1) and 3), it seems to be due to the similarities of the denominator in the original integral...I've never had to deal with a denominator with two different irreducible fractions.

If I did anything wrong, please point it out, or enlighten me toward the appropriate approach.

 

[dextercioby]

Let's stick to the original "t",so the OP could make the connection with the integral which started the thread.

$$\frac{At+B}{1+t\sqrt{2}+t^2}+\frac{Ct+D}{1-t\sqrt{2}+t^2}\equiv \frac{t^{2}}{\mbox{product}}$$

Cross multiplying,one gets the system of equations

$$\left\{\begin{array}{c} A+C=0\\B+D+C\sqrt{2}-A\sqrt{2}=1\\ A+C+D\sqrt{2}-B\sqrt{2}=0\\B+D=0 \end{array}\right$$

which has the unique solution

$$\left\{\begin{array}{cccc} A=-\frac{\sqrt{2}}{4}, & B=0, & C=\frac{\sqrt{2}}{4}, & D=0 \end{array} \right\}$$

Can u carry on from here...?

Daniel.

 

[whozum]

Can you show where GCT went wrong? I followed his procedure and can't spot an error.

 

[dextercioby]

H mixed "t" and "x"...Not to mention that the system of equations should have been variable ('x' or 't' or whatever)-free.

Daniel.

 

[whozum]

I spotted the x,t switch but I figured it was a notation issue. Thanks.

 

[GCT]

Man...this is not my day

anyways

$$\left\{\begin{array}{c} A+C=0\\B+D+C\sqrt{2}-A\sqrt{2}=1\\ A+C+D\sqrt{2}-B\sqrt{2}=0\\B+D=0 \end{array}\right$$

the third and last equation seems to contradict each other, substitute A+C=0 into the third, indicates B=D, the fourth equation indicates B=-D.

 

[whozum]

B and D are zero as dex pointed out.

 

[GCT]

okay, I worked it out and the solution seems to be similar to #33, an innocent looking integral...turned out to be very nasty in the end.

 

[GCT]

here's my work

$$I=2 \int \frac{- \sqrt{2} t/4}{1+ \sqrt{2}t +t^{2}}~+~ \frac{ \sqrt{2} t/4}{1- \sqrt{2}t +t^{2}}~dt$$

solving for the first component

$$\int \frac{(- \sqrt{2}/8)( \sqrt{2} +2t_)}{1+ \sqrt{2}t +t^{2}}~dt~+ <br /> (1/4) \int \frac{1}{1+ \sqrt{2} t +t^{2}}~dt$$


$$u=1+ \sqrt{2} t + t^{2},~du= \sqrt{2} + 2t~dt$$

$$(- \sqrt{2} / 8) ln(1+ \sqrt{2} t + t^{2}) + ( \sqrt{2} /4) arctan( \sqrt{2} + 1))$$

the second component turns out to be
$$( \sqrt{2} / 8) ln (1- \sqrt{2} t + t^{2}) + ( \sqrt{2} /4) <br /> arctan( \sqrt{2} - 1)$$

adding them up, in nonsimplified form

$$(- \sqrt{2} / 4) {ln ( \frac {1- \sqrt{2} \sqrt{tanx} + tanx}{1+ \sqrt{2} \sqrt {tanx} + tanx})} + ( \sqrt{2} /2) {(arctan( \sqrt{2} + 1) + arctan( \sqrt{2} - 1)} +C$$

didn't know the absolute value latex, see any mistakes...let me know

 

[whozum]

Can you elaborate on what you did in the first two latex expressions?

 

[GCT]

well

$$\frac{- \sqrt{2} t/4}{1+ \sqrt{2}t +t^{2}}$$=

$$\frac{(- \sqrt{2}/8)( \sqrt{2} +2t_) + (1/4)}{1+ \sqrt{2}t +t^{2}}~dt~$$

Since

$$u=1+ \sqrt{2} t + t^{2},~du= \sqrt{2} + 2t~dt$$

you'll first need to find this derivative and sort of work backwards. note that

$$- \sqrt{2} t/4~=~- \sqrt{2}/8)( \sqrt{2} +2t_) + (1/4)$$

 

[whozum]

$$- \sqrt{2} t/4~=~- \sqrt{2}/8)( \sqrt{2} +2t_) + (1/4)$$

How did you find this expression? Is there a need to do so?

 

[GCT]

It somewhat of a trick I learned in integral calculus this past semester. You'll notice that one of the fractions contains a derivative of the denominator,

First we note that the derivative of

$$1+ \sqrt{2}t +t^{2}$$

is $$\sqrt{2} + 2t$$

We'll try to arrive at form where we can separate the fraction into two simpler fractions.

$$\frac{(- \sqrt{2} t/4)}{ 1+ \sqrt{2}t +t^{2}}$$

1)$$\frac{ \sqrt{2} + 2t}{1+ \sqrt{2}t +t^{2}}$$

we'll try resolving the numerator

2)$$\frac{( \sqrt{2}/8)( \sqrt{2} +2t)}{1+ \sqrt{2}t +t^{2}}$$

we now have an equivalent "t" coefficient, we'll need to resolve the nonvariable component, first find the nonvariable component turns out to be $(- \sqrt{2}/8)*( \sqrt{2})~=~-1/4$ thus we'll need to add $1/4$

3)$$\frac{(- \sqrt{2}/8)( \sqrt{2} +2t_) + (1/4)}{1+ \sqrt{2}t +t^{2}}~dt~$$

which is equivalent to the original form...now you can separate it into two simpler components. Integrate the first using substitution and the second by converting the denominator into a actangent derivative.