Why Are My Tangent Half-Angle Substitution Answers Different from the Workbook?

[Dethrone]

I did two questions from my workbook that involved the tangent half-angle substitution, \(\displaystyle z = tan (\frac{x}{2})\). The answers that I got, for two questions, were different (but correct, I think) in the same way. Can you assist me in how to acquire the workbook answer?

1. \(\displaystyle \int \frac{dx}{1-2sinx} \)

Workbook answer:
\(\displaystyle \int \frac{\sqrt{3}}{3} \ln\left({\frac{\tan\left({\frac{x}{2}}\right)-2-\sqrt{3}}{\tan\left({\frac{x}{2}}\right)-2+\sqrt{3}}}\right)\)

My answer:
\(\displaystyle \int \frac{2\sqrt{3}}{3}\ln\left({\frac{\tan\left({\frac{x}{2}}\right)-2-\sqrt{3}}{\sqrt{(\tan\left({\frac{x}{2}}\right)-2)^2-3}}}\right)\)

Method:
Applying the tangent half-angle substitution and completing the square in the denominator, I get
\(\displaystyle \int \frac{2}{(z-2)^2-3} dz\)

Letting \(\displaystyle z - 2 = u = \sqrt{3}sec\theta\), I end up with
\(\displaystyle \frac{2\sqrt{3}}{3} \int csc\theta d\theta\)

Evaluating that and putting subbing back all the variables gives me the answer I wrote above. The same method was applied in number 2.
2. \(\displaystyle \int \frac{sinx}{1+sin^2x} dx\)

Workbook answer:
\(\displaystyle \frac{\sqrt{2}}{4}\ln\left({\frac{(\tan\left({\frac{x}{2}}\right))^2+3-2\sqrt{2}}{(\tan\left({\frac{x}{2}}\right))^2+3+2\sqrt{2}}}\right)\)

My answer:

\(\displaystyle \frac{\sqrt{2}}{2}\ln\left({\frac{(\tan\left({\frac{x}{2}}\right))^2+3-\sqrt{8}}{\sqrt{((\tan\left({\frac{x}{2}}\right))^2+3)^2-8}}}\right)\)

As is seen, my answer differs from the workbook in the same manner for both questions. I tried using wolfram alpha, which obtains the same answer as the workbook, but it uses the tanh substitution, which I don't think was used in the workbook. Which integration method did the textbook use? (I have come to love LaTex, especially with all the user-friendly commands on the side :) )

 

[Dethrone]

I was scouring the internet and I found what they did.

\(\displaystyle \int \frac{2}{(z-2)^2-3} dz\)

At this step, I generally use a secant substitution; wouldn't that be the most obvious choice anyways? Well, they used a difference of squares factoring in the denominator and applied partial fraction decomposition.

\(\displaystyle \int \frac{2}{(z-2-\sqrt{3})(z-2+\sqrt{3})} dz\)

 

[I like Serena]

Hi Rido! ;)

All 3 methods are equally valid.

However, in both problems your answer can be simplified further.
If you do, you'll get the textbook answer.

 

[Dethrone]

I tried to simplify it before with limited success. Any hints? Let's try 1)

\(\displaystyle \displaystyle \int \frac{2\sqrt{3}}{3}\ln\left({\frac{\tan\left({\frac{x}{2}}\right)-2-\sqrt{3}}{\sqrt{(\tan\left({\frac{x}{2}}\right)-2)^2-3}}}\right)\)

 

[I like Serena]

I tried to simplify it before with limited success. Any hints? Let's try 1)

\(\displaystyle \displaystyle \int \frac{2\sqrt{3}}{3}\ln\left({\frac{\tan\left({\frac{x}{2}}\right)-2-\sqrt{3}}{\sqrt{(\tan\left({\frac{x}{2}}\right)-2)^2-3}}}\right)\)

Oh wait. There is something wrong after all.
Apparently there shouldn't be a radical in the denominator.
Then it would simplify properly.

Edit: Nope. It's right after all.

 

[Dethrone]

That's weird. My substitution is that \(\displaystyle z-2=u=\sqrt{3}sec\theta\)

Constructing a right angle triangle, I get that the hypotenuse is u, adjacent is \(\displaystyle \sqrt{3}\), and opposite is \(\displaystyle \sqrt{u^2-3}\)

From \(\displaystyle \displaystyle \frac{2\sqrt{3}}{3} \int csc\theta d\theta\) (in my first post)

I get \(\displaystyle \frac{2\sqrt{3}}{3}\ln\left({csc\theta-cot\theta}\right)\)

Replacing the trig functions and letting u = z-2, and z = tan(x/2), I'm pretty I get

\(\displaystyle \displaystyle \displaystyle \int \frac{2\sqrt{3}}{3}\ln\left({\frac{\tan\left({\frac{x}{2}}\right)-2-\sqrt{3}}{\sqrt{(\tan\left({\frac{x}{2}}\right)-2)^2-3}}}\right) \)

 

[I like Serena]

Sorry for the confusion.
It does simplify properly.
$$\ln\left({\frac{\tan\left({\frac{x}{2}}\right)-2-\sqrt{3}}{\sqrt{(\tan\left({\frac{x}{2}}\right)-2)^2-3}}}\right)

= \ln\left(\sqrt{{\frac{\left((\tan\left({\frac{x}{2}}\right)-2)-\sqrt{3}\right)^2}{(\tan\left({\frac{x}{2}}\right)-2)^2-3}}}\right)

= \frac 1 2 \ln\left({\frac{(\tan\left({\frac{x}{2}}\right)-2)-\sqrt{3}}{(\tan\left({\frac{x}{2}}\right)-2)+\sqrt 3}}\right)
$$

 

[I like Serena]

Btw, did you know that:

$$\int \frac{dx}{1 - x^2} = \text{artanh }x + C$$

And that:
$$\text{artanh }x = \frac 1 2 \ln \left( \frac{1+x}{1-x} \right)$$

 

[Dethrone]

Well, thanks for all the help! I guess this question is now official solved. :)
Wolfram Alpha showed me that, but I haven't learned it yet. Heck, I just self-learned integration so far, so I really appreciate the help :)