Factorizing denominator help

Yegor

$$\int\frac{1}{(\sin(x)+\frac{2}{\cos(x)})^2} dx$$
I know that tan(x)=t works, but there are some problems with factorizing denominator and even after factorizing that integral isn't very simple. Does anyone see here any simplier substitution? thank you

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maverick280857

Its $tan(\frac{x}{2}) = t$. Try it. What are the problems?

dextercioby

Homework Helper
Don't be afraid of this simple algebra.When doing integrals,you ought to be prepared for worst.

BTW,LaTex acknowledges the function "tangent",as long as you write it " \tan ".

Daniel.

Yegor

In this case denominator is $$t^4-t^3+2t^2+t+1$$. This i can't factorize also. Even Mathematica doesn't help.

Yegor

Hm. Yes, i'm probably just too lazy today to work on algebra. But i have one more question about such problems. If i have polynomials with complex solutions what is the algorithm for factorizing them?
As examples:1) $$t^4-t^3+2t^2+t+1$$
2) $$4t^4+4t^3+9t^2+4t+4$$

dextercioby

Homework Helper
It can't be.

$$I= \int \frac{dx}{\left(\sin x+\frac{2}{\cos x}\right)^{2}}$$ (1)

Make the sub

$$\tan\frac{x}{2}=t$$ (2)

An you'll get the integral

$$I=2\int \frac{\cos ^2\left( 2\arctan t\right) }{\left[ \sin \left( 2\arctan t\right) \cos \left( 2\arctan t\right) +2\right] ^2\left( 1+t^2\right) }\,dt$$ (3)

Then

$$\left\{\begin{array}{c} \cos \left( \arctan t\right) =\frac{1}{\sqrt{\left( 1+t^2\right) }}\\ \sin \left( \arctan t\right) =\frac {t}{\sqrt{\left( 1+t^2\right) }} \end{array} \right$$ (4)

Can you take it from here...?

Daniel.

Last edited:

maverick280857

dextercioby said:
Don't be afraid of this simple algebra.When doing integrals,you ought to be prepared for worst.

BTW,LaTex acknowledges the function "tangent",as long as you write it " \tan ".

Daniel.
Yeah I forgot. Thanks anyway.

Yegor

$$\left{\begin{array}{c} \cos \left( 2\arctan t\right) =\frac{1-t^2}{{\left( 1+t^2\right) }}\\ \sin \left( 2\arctan t\right) =\frac {2t}{{\left( 1+t^2\right) }} \end{array} \right$$

Yegor

Yes, i did it dextercioby. This i have from it:
$$\int\frac{(1+t^2)(1-t^2)^2}{t^4-t^3+2t^2+t+1}$$
OK. i can devide this fraction, but i can do nothing with denominator

dextercioby

Homework Helper
Here's what i'd do.I denote your integral by "I".

$$I=:\int\frac{\cos^{2}x}{\left(\frac{1}{2}\sin 2x +2\right)^{2}} \ dx$$ (1)

I define its sister integral

$$J=:\int\frac{\sin^{2}x}{\left(\frac{1}{2}\sin 2x +2\right)^{2}} \ dx$$ (2)

The difference

$$I-J=\int \frac{\cos 2x}{\left(\frac{1}{2}\sin 2x +2\right)^{2}} \ dx =\int \frac{d\left(\frac{1}{2}\sin 2x +2\right)}{\left(\frac{1}{2}\sin 2x +2\right)^{2}} =-\frac{1}{\frac{1}{2}\sin 2x +2}$$ (3)

I'll concentrate on the sum.

To be continued.

Daniel.

dextercioby

Homework Helper
Here's the sum

$$I+J=\int \frac{dx}{\left(\frac{1}{2}\sin 2t +2\right)^2}$$ (4)

I make the sub

$$2x=t$$ (5)

$$I+J=\frac{1}{2}\int \frac{dt}{\left(\frac{1}{2}\sin t +2\right)^{2}}$$ (6)

I make the sub

$$\tan\frac{t}{2}=u$$ (7)

$$I+J=\int \frac{1+u^{2}}{(2u^{2}+u+2)^{2}} \ du=\frac{1}{4}\int \frac{1+u^{2}}{\left[\left(u+\frac{1}{4}-\frac{i\sqrt{3}}{2}\right)\left(u+\frac{1}{4}+\frac{i\sqrt{3}}{2}\right)\right]^{2}} \ du$$ (8)

Therefore

$$I+J=\int \frac{1+u^{2}}{\left[u^{2}+u\left(\frac{1}{2}-i\sqrt{3}\right)-\frac{11}{16}-\frac{i\sqrt{3}}{4}\right]\left[u^{2}+u\left(\frac{1}{2}+i\sqrt{3}\right)-\frac{11}{16}+\frac{i\sqrt{3}}{4}\right]} \ du$$ (9)

and now apply simple fractions.

Daniel.

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