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If $\int \frac{d x}{\cos^{6} x + \sin^{6} x} = \tan^{- 1} (k \tan 2 x) + C$ , then

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a

k=1/2

b

k=1/4

c

k=1/6

d

k depends on x

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detailed solution

Correct option is A

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.

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If $\int \frac{s i n^{2} x}{1 + s i n^{2} x} d x = x - K t a n^{- 1} (M t a n x) + C$ then

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