Integration Question?

Question

dy/dx= (cos x -sin x)^2

can I do that sum this way:

y= (cosx-sinx)^3 / 3(sinx + cosx)

care to tell me why not?

Answer 1

( cos x - sin x ) ²
= cos ² x - sin ² x - 2 sin x cos x
= cos 2x - sin 2x
I = ∫ cos 2x dx - ∫ sin 2x dx
I = Ia + Ib
I = (1/2) sin 2x + (1/2) cos 2x + C

Note
Ia and Ib are standard integrals
∫ (cos x - sin x) dx is not a standard integral.

Answer 2

sorry, but no.... to get ( cos x - sin x )^3, I assume you tried U = cos x - sin x , but then you have
dy / dx = U^2 , but to integrate, you need dy / dU , and I don't see any transformation you can try that will be valid....

here's what I would do...

(cos x - sin x )^2 =
(cos x )^2 - 2sin x cos x + ( sin x )^2
by trig id.... [cos x]^2 = 0.5 [ 1 + cos(2x) ] ,
and [sin x ]^2 = 0.5 [ 1 - cos (2x) ]

= 0.5 + 0.5 cos(2x) - 2sin x cos x + 0.5 - 0.5 cos(2x) =
1 - 2 sin(x) cos(x)

easy to integrate 1 - 2 sin(x) cos(x) from here ...
for the second term, try U = sin (x) as a substitution...