Answer to Question #193038 in Linear Algebra for Regomoditswe Dibob

Question

$Cosy dy/dx$ = $Sin^2x Cosx$

Answer 1

Given,

$cosy\cdot \dfrac{dy}{dx}=sin^2x\times cosx\\\ \\cosy\times dy=sin^2x\times cosxdx$

Now, Integrating both sides

$\int cosydy=\int sin^2xcosxdx$

$Let \ I_1=\int sin^2xcosxdx\\$

Substitute sinx = t

then cosxdx = dt

$I_1=\int t^2dt\\I_1=\frac{t^3}{3}\\$

Now, $I_1= \dfrac{1}{3}sin^3x$

So,

$\int cosydy=\int sin^2xcosxdx$

$siny =\dfrac{1}{3}sin^3x +C\\\boxed{y=sin^{-1}(\frac{1}{3}sin^3x+C)}$