How Do You Integrate the Square Root of Tangent?

[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

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


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

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

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

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.