twoflower
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Hi,
I've been having troubles solving this integral:
<br /> \int \frac{\sin x.\cos x}{\sin^4 x + x.\cos^4 x} dx<br />
Here's how I tried it:
<br /> t = \cos x <br />
<br /> dt = - \sin x dx<br />
<br /> dx = \frac{dt}{-\sin x}<br />
<br /> x = \arccos t<br />
<br /> dx = -\frac{1}{\sqrt{1-t^2}} dt<br />
<br /> -\frac{1}{\sin x} = -\frac{1}{\sqrt{1-t^2}}<br />
<br /> \sin x = \sqrt{1-t^2}<br />
<br /> \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<br />
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:
<br /> \int \frac{\sin x.\cos x}{\sin^4 x + x.\cos^4 x} dx<br />
Here's how I tried it:
<br /> t = \cos x <br />
<br /> dt = - \sin x dx<br />
<br /> dx = \frac{dt}{-\sin x}<br />
<br /> x = \arccos t<br />
<br /> dx = -\frac{1}{\sqrt{1-t^2}} dt<br />
<br /> -\frac{1}{\sin x} = -\frac{1}{\sqrt{1-t^2}}<br />
<br /> \sin x = \sqrt{1-t^2}<br />
<br /> \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<br />
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?