Evaluate the following integrals:

Assume acos²x + bsin²x = t

d(acos²x + bsin²x) = dt

( - 2acosx.sinx + 2bsinx.cosx)dx = dt

(bsin2x - asin2x)dx=dt

(b - a)sin2xdx=dt

Sin2xdx =

Put t and dt in given equation we get

⇒

= ln|t| + c.

But t = acos²x + bsin²x

= ln| acos²x + bsin²x | + c.