- #1
twoflower
- 368
- 0
Hi,
I've been having troubles solving this integral:
[tex]
\int \frac{\sin x.\cos x}{\sin^4 x + x.\cos^4 x} dx
[/tex]
Here's how I tried it:
[tex]
t = \cos x
[/tex]
[tex]
dt = - \sin x dx
[/tex]
[tex]
dx = \frac{dt}{-\sin x}
[/tex]
[tex]
x = \arccos t
[/tex]
[tex]
dx = -\frac{1}{\sqrt{1-t^2}} dt
[/tex]
[tex]
-\frac{1}{\sin x} = -\frac{1}{\sqrt{1-t^2}}
[/tex]
[tex]
\sin x = \sqrt{1-t^2}
[/tex]
[tex]
\int \frac{\sin x.\cos x}{\sin^4 x + x.\cos^4 x} dx = - \int \frac{\cos x}{\sin^4 x + x.\cos^4 x} . \left( -\sin x \right) dx = - \int \frac{t}{\left(1-t^2\right)^2 + t^4.\arccos t} dt
[/tex]
Well, I don't know at all what to do with this now...
Does anybody have any idea?
PS: Why doesn't it insert newline, when I write "\\" in the LaTeX code?
I've been having troubles solving this integral:
[tex]
\int \frac{\sin x.\cos x}{\sin^4 x + x.\cos^4 x} dx
[/tex]
Here's how I tried it:
[tex]
t = \cos x
[/tex]
[tex]
dt = - \sin x dx
[/tex]
[tex]
dx = \frac{dt}{-\sin x}
[/tex]
[tex]
x = \arccos t
[/tex]
[tex]
dx = -\frac{1}{\sqrt{1-t^2}} dt
[/tex]
[tex]
-\frac{1}{\sin x} = -\frac{1}{\sqrt{1-t^2}}
[/tex]
[tex]
\sin x = \sqrt{1-t^2}
[/tex]
[tex]
\int \frac{\sin x.\cos x}{\sin^4 x + x.\cos^4 x} dx = - \int \frac{\cos x}{\sin^4 x + x.\cos^4 x} . \left( -\sin x \right) dx = - \int \frac{t}{\left(1-t^2\right)^2 + t^4.\arccos t} dt
[/tex]
Well, I don't know at all what to do with this now...
Does anybody have any idea?
PS: Why doesn't it insert newline, when I write "\\" in the LaTeX code?