∫π2ocos2xdxcos2x+4sin2x
=∫π2ocos2xdxcos2x+4(1−cos2x)
=∫π2ocos2xdx−3cos2x+4
=−13∫π2o(−3cos2x+4)−4dx−3cos2x+4
=−13∫π2o(1−4dx4−3cos2x)
=−13[x]π20+43∫π2o(dx4−3cos2x)
=−π6+43∫π2o(sec2xdx4sec2x−3)
Dividing Nr and Dr by cos2x
=−π6+43∫π2o(sec2xdx4(1+tan2x)−3)
put tanx=t
sec2xdx=dt
=−π6+43∫∞o(dt4(1+t2)−3)
=−π6+43∫∞0(dt2t2+1)
=−π6+43×12[tan−1(2t)]∞0=π3