Let I = \(\int\limits_{0}^{\pi/2}\sqrt{cos\,\text x-cos^3\text x}(sec^2\text x-1) \)cos2x dx
We have sin2x + cos2x = 1 and sec2x – tan2x = 1
We can write sin3x = sin2x \(\times\) sin x = (1 – cos2x) sin x
Put cos x = t
⇒ –sin(x)dx = dt (Differentiating both sides)
⇒ sin(x)dx = –dt
When x = 0, t = cos 0 = 1
When x = \(\cfrac{\pi}2\), t = cos \(\cfrac{\pi}2\) = 0
So, the new limits are 1 and 0.
Substituting this in the original integral,