# Thread: I forgot the clever way of doing these integrals...

1. ## I forgot the clever way of doing these integrals...

$\int \frac{\sin^2 x - 4\sin x \cos x + 3\cos^2 x}{\sin x +\cos x}dx$

2. ## Re: I forgot the clever way of doing these integrals...

Originally Posted by leehuan
$\int \frac{\sin^2 x - 4\sin x \cos x + 3\cos^2 x}{\sin x +\cos x}dx$
$\noindent 4 + \sin^2{x} - 4\sin{x}\cos{x} + 3\cos^2{x} - 4 \equiv 4 - 3\sin^2{x} - 4\sin{x}\cos{x} - \cos^2{x} \\\\ \equiv 4-(\sin{x}+\cos{x})(3\sin{x}+\cos{x})$

3. ## Re: I forgot the clever way of doing these integrals...

Originally Posted by leehuan
$\int \frac{\sin^2 x - 4\sin x \cos x + 3\cos^2 x}{\sin x +\cos x}dx$
If numerator were: sin2x + 4sinxcosx + 3cos2x things would be much easier.

In this case numerator = (sinx + 3cosx)(sinx + cosx). But as it stands, I have not figured out how. Alternatively, were the denominator (sinx - cos x) it'd also be simple.

The numerator as it stands = (sinx - cosx)(sinx - 3cosx)

4. ## Re: I forgot the clever way of doing these integrals...

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