MHB Solving Difficult Integral: $\int \frac{dx}{1+\sin(x) -\cos^2(x)}$

[leprofece]

View attachment 2974
Viewing the image
the integral is
$$\int \frac{dx}{1+\sin(x) -\cos^2(x)}$$

I am going to prove mathtype this time
View attachment 2975
Well I am getting better
I put or use the theorem of the image and I put
$$\frac{dx}{\sin^2(x)+\sin(x)} $$
I substitute and I got a difficult unsolvable integral

 

[I like Serena]

View attachment 2974
Viewing the image
the integral is
$$\int \frac{dx}{1+\sin(x) -\cos^2(x)}$$

Hi leprofece!

As the first step, let's do only the substitution.
$$
\int \frac{dx}{1+\sin x - \cos^2 x}
= \int \frac{1}{1+\frac{2z}{1+z^2}-\left(\frac{1-z^2}{1+z^2}\right)^{\!2}} \cdot \frac{2dz}{1+z^2}
$$

Can you simplify it?

 

[chisigma]

View attachment 2974
Viewing the image
the integral is
$$\int \frac{dx}{1+\sin(x) -\cos^2(x)}$$

I am going to prove mathtype this time
View attachment 2975
Well I am getting better
I put or use the theorem of the image and I put
$$\frac{dx}{\sin^2(x)+\sin(x)} $$
I substitute and I got a difficult unsolvable integral

Why don't proceed step by step?... first step: the integral can be broken into two more simple integrals because is...

$\displaystyle \frac{1}{\sin^{2} x + \sin x} = \frac{1}{\sin x} - \frac{1}{1 + \sin x}\ (1)$

Second step : now You can apply the suggested substitution to both terms of (1)...

Are You able to proceed?... Kind regards

$\chi$ $\sigma$

 

[leprofece]

this is my work and my answer am I correct?
View attachment 2994

 

[Dethrone]

Your work is a bit messy, so I can't tell where you went wrong.

Starting from where ILS left off:

$$=\int \frac{1}{1+\frac{2z}{1+z^2}-\left(\frac{1-z^2}{1+z^2}\right)^{\!2}} \cdot \frac{2dz}{1+z^2}$$

Simplifying it:

$$=\int \frac{2}{\left(\frac{1+z^2+2z}{1+z^2}-\frac{(1-z^2)^2}{(1+z^2)^2}\right)(1+z^2)}dz$$
$$=\int \frac{2}{\frac{(1+z^2+2z)(1+z^2)-(1-z^2)^2}{1+z^2}}dz$$

Do you see your mistake?

 

[leprofece]

well after the mistake I got
View attachment 3003

these lnx+xlnx ... is the integral answer

 

[Dethrone]

You are right up to:

$$\int \frac{1+z^2}{z(1+z)^2}dz$$

Can you explain how you got to the next step? Did you use partial fractions?