twoflower
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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?