∫tan⁡xsin4⁡x+cos4⁡xdx is equal to

LetI=∫tan⁡xsin4⁡x+cos4⁡xdx= l=∫tan⁡xsec2⁡xtan4⁡x+1dx=12∫1tan2⁡x2+1dtan2⁡x⇒ I=12loge⁡tan2⁡x+1+tan4⁡x+C

∫tan⁡xsin4⁡x+cos4⁡xdx is equal to

LetI=∫tan⁡xsin4⁡x+cos4⁡xdx= l=∫tan⁡xsec2⁡xtan4⁡x+1dx=12∫1tan2⁡x2+1dtan2⁡x⇒ I=12loge⁡tan2⁡x+1+tan4⁡x+C

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$\int \frac{\tan x}{\sqrt{\sin^{4} x + \cos^{4} x}} d x$ is equal to

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$\log_{e} (\tan^{2} x + \sqrt{1 + \tan^{4} x}) + C$

$\frac{1}{2} \log_{e} (\tan^{2} x + \sqrt{1 + \tan^{4} x}) + C$

none of these

$\frac{1}{4} \log (\tan^{2} x + \sqrt{1 + \tan^{4} x}) + C$

answer is B.

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Detailed Solution

Let

$\begin{array}{l} I = \int \frac{\tan x}{\sqrt{\sin^{4} x + \cos^{4} x}} d x \\ = l = \int \frac{\tan x \sec^{2} x}{\sqrt{\tan^{4} x + 1}} d x = \frac{1}{2} \int \frac{1}{\sqrt{{(\tan^{2} x)}^{2} + 1}} d (\tan^{2} x) \\ \Rightarrow I = \frac{1}{2} \log_{e} \{\tan^{2} x + \sqrt{1 + \tan^{4} x} + C \end{array}$