If ∫dxcos6x+sin6x=tan−1(ktan2x)+C, then
k=1/2
k=1/4
k=1/6
k depends on x
1cos6x+sin6x=11−3sin2xcos2xsec4xsec4x−3tan2x=1+tan2xsec2x1+tan2x2−3tan2x
Putting tanx=t the given integral reduces to
∫1+t21+t4−t2dt=∫1+1/t2t2+1/t2−1dt=∫duu2+1,u=t−1/t=tan−1(tanx−cotx)+C=cot−1−12tan2x+C=tan−112tan2x+ Const.