Assume x + cos²x =t

d(x + cos²x) =dt

(1 + ( - 2cosx.sinx))dx = dt

2sinx.cosx=sin2x

(1 - sin2x)dx=dt

Put t and dt in given equation we get

⇒

=

But t = x + cos²x

= ln| x + cos²x | + c.