I forgot the clever way of doing these integrals... (1 Viewer)

Paradoxica · Jan 24, 2018

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

$$\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})$

Drongoski · Jan 24, 2018

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

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)

Paradoxica · Feb 8, 2018

http://community.boredofstudies.org...egration-marathon-archive-14.html#post7092990

I forgot the clever way of doing these integrals... (1 Viewer)

leehuan

Well-Known Member

Paradoxica

-insert title here-

Drongoski

Well-Known Member

Paradoxica

-insert title here-

Users Who Are Viewing This Thread (Users: 0, Guests: 1)