If ∫dxcos6⁡x+sin6⁡x=tan−1⁡(ktan⁡2x)+C, then

1cos6⁡x+sin6⁡x=11−3sin2⁡xcos2⁡xsec4⁡xsec4⁡x−3tan2⁡x=1+tan2⁡xsec2⁡x1+tan2⁡x2−3tan2⁡xPutting tan⁡x=t the given integral reduces to∫1+t21+t4−t2dt=∫1+1/t2t2+1/t2−1dt=∫duu2+1,u=t−1/t=tan−1⁡(tan⁡x−cot⁡x)+C=cot−1⁡−12tan⁡2x+C=tan−1⁡12tan⁡2x+ Const.