MHB Solving a Complex Integral: Substituting tg (x/2)

[leprofece]

integral View attachment 2995

This is my work but I got a very looong integral to solve
after substitute tg (x/2) based in my former exercise

View attachment 2996
it remains 4/ (1-z^2)(3-z^2)

 

[Prove It]

integral View attachment 2995

This is my work but I got a very looong integral to solve
after substitute tg (x/2) based in my former exercise

https://www.physicsforums.com/attachments/2996
it remains 4/ (1-z^2)(3-z^2)

Each of those can be factorised further and you can apply partial fractions.

However I think it might be possible to avoid the Wierstrauss Substitution...

$\displaystyle \begin{align*} \int{ \frac{1 + \cos{(x)}}{\cos{(x)} \left[ 1 + 2\cos{(x)} \right] } \, \mathrm{d}x} &= \int{ \frac{ \left[ 1 + \cos{(x)} \right] \left[ 1 - \cos{(x)} \right] }{ \cos{(x)} \left[ 1 + 2\cos{(x)} \right] \left[ 1 - \cos{(x)} \right] } \,\mathrm{d}x} \\ &= \int{ \frac{1 - \cos^2{(x)}}{\cos{(x)} \left[ 1 + 2\cos{(x)} \right] \left[ 1 - \cos{(x)} \right] } \, \mathrm{d}x } \\ &= \int{ \frac{\sin^2{(x)} }{ \cos{(x)} \left[ 1 + 2\cos{(x)} \right] \left[ 1 - \cos{(x)} \right] } \,\mathrm{d}x} \\ &= -\int{ \frac{\sqrt{ 1 - \cos^2{(x)} } \left[ - \sin{(x)} \right] }{ \cos{(x)} \left[ 1 + 2\cos{(x)} \right] \left[ 1 - \cos{(x)} \right] } \, \mathrm{d}x } \end{align*}$

Now you could try the substitution $\displaystyle \begin{align*} u = \cos{(x)} \implies \mathrm{d}u = -\sin{(x)} \,\mathrm{d}x \end{align*}$...

 

[Dethrone]

Since you're at the home stretch, elaborating on what Prove It said:

$$=\int \frac{4}{(1-z^2)(3-z^2)} \,dz$$
$$=\int \frac{4}{(z^2-1)(z^2-3)}\,dz$$
$$=\int \frac{4}{(z+1)(z-1)(z^2-3)}\,dz$$

Now apply partial fractions. :D

 

[Prove It]

Since you're at the home stretch, elaborating on what Prove It said:

$$=\int \frac{4}{(1-z^2)(3-z^2)} \,dz$$
$$=\int \frac{4}{(z^2-1)(z^2-3)}\,dz$$
$$=\int \frac{4}{(z+1)(z-1)(z^2-3)}\,dz$$

Now apply partial fractions. :D

$\displaystyle \begin{align*} &= \int{ \frac{4}{ \left( z + 1 \right) \left( z - 1 \right) \left( z + \sqrt{3} \right) \left( z - \sqrt{3} \right) } \, \mathrm{d}z} \end{align*}$

 

[leprofece]

Ok let's see
this integral is too long according to wolphramalpha
maybe as prove it did it, it is easier
thanks anyway

 

[Dethrone]

I'm not Wolfram Alpha, nor a CAS, but the integral does seem manageable.

$$=\int \frac{4}{(z^2-1)(z^2-3)}\,dz$$

I was hesitant to factor the $z^2-3$ because it has a known antiderivative, and if you know what that is, it would be easier to split the integral as

$$=\int \frac{4}{(z^2-1)(z^2-3)}\,dz$$

Otherwise, if you want to further pursue this problem, work from and here and try partial fractions.

\begin{align*} &= \int{ \frac{4}{ \left( z + 1 \right) \left( z - 1 \right) \left( z + \sqrt{3} \right) \left( z - \sqrt{3} \right) } \, \mathrm{d}z} \end{align*}