∫cos2x−cos2acosx−cosadx
We know that cos2x=2cos2x−1
=∫2cos2x−1−2cos2a+1cosx−cosadx
=2∫cos2x−cos2acosx−cosadx
=2∫(cosx−cosa)(cosx+cosa)cosx−cosadx
=2∫(cosx+cosa)dx
=2(sinx+xcosa)+C
Evaluate limx→0(1−cos x√cos2x)x2