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

A
$k = 1 / 2$
B
$k = 1 / 4$
C
$k = 1 / 6$
D
$k$ depends on $x$

Solution:

$\begin{array}{l} \frac{1}{\cos^{6} x + \sin^{6} x} = \frac{1}{1 - 3 \sin^{2} x \cos^{2} x} \\ \frac{\sec^{4} x}{\sec^{4} x - 3 \tan^{2} x} = \frac{(1 + \tan^{2} x) \sec^{2} x}{{(1 + \tan^{2} x)}^{2} - 3 \tan^{2} x} \end{array}$

Putting $\tan x = t$ the given integral reduces to

$\begin{array}{l} \int \frac{1 + t^{2}}{1 + t^{4} - t^{2}} d t = \int \frac{1 + 1 / t^{2}}{t^{2} + 1 / t^{2} - 1} d t \\ = \int \frac{d u}{u^{2} + 1}, u = t - 1 / t \\ = \tan^{- 1} (\tan x - \cot x) + C \\ = \cot^{- 1} (- \frac{1}{2} \tan 2 x) + C \\ = \tan^{- 1} (\frac{1}{2} \tan 2 x) + Const. \end{array}$

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

Solution:

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