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.