# [SOLVED] integrating factor gone wrong

• May 29th 2009, 04:02 PM
i_zz_y_ill
[SOLVED] integrating factor gone wrong
ok so i dont get this question, the integrating factor is clearly sinx from which point i decided that if you minus each side you can abreviate it to $\frac{d(-sinxf)}{dx}=-cos^2xsinx$ NB original question is: $\frac{df}{dx}+fcotx=cos^2x$ the answer is somehow $f(x)=-\frac{cos^3x}{3sinx}+\frac{C}{sinx}$ which i do no understand,,,how can the constand have a sinx underneath is isnt this just another constant? I must be going wrong somewhere on the integral of $\int cos^2dx$ which i make out to be,,, $\frac{co3^3x}{3}$ ? any clues?
• May 29th 2009, 04:17 PM
Spec
$\frac{d}{dx}\left(\frac{1}{3}\cos^3 x \right)\neq \cos^2 x$
• May 29th 2009, 04:29 PM
i_zz_y_ill
oh yeah! now using $\int \frac{1}{2}+\frac{cos2x}{2}dx$= ???
• May 29th 2009, 04:40 PM
Random Variable
$sin(x) \frac {df}{dx} + f sin(x) \frac {cosx}{sinx} = sin(x)cos^{2}x$

$\int \frac {d}{dx} (fsin(x))dx = \int sin(x)cos^{2}(x)dx$

$fsin(x) = - \int u^2 du = -\frac {1}{3}u^{3} + C = - \frac {1}{3}cos^{3}x + C$

then solve for f
• May 29th 2009, 05:21 PM
i_zz_y_ill
what is u? still bit unsure how to solve that last integral? $\int \frac{sinx}{2}+\int sinx -2 \int sin^3x ?$

$sinxcos^2x=sinx(\frac{1}{2}+cos2x)=\frac{sinx}{2}+ sinx(1-2sin^2x)$?

any ideas? (Lipssealed)
• May 29th 2009, 05:23 PM
Spec
$\int \sin x \cos^2 x dx =\left\{u=\cos x, du=-\sin x dx \implies dx=-\frac{du}{\sin x}\right\}=\int -u^2 du$