∫e2log⁡(tan⁡x)−e2log⁡(cot⁡x)cos2⁡xdx=f(x)+C then fπ4=

I=∫elog⁡tan2⁡x−elog⁡cot2⁡xcos2⁡xdxI=∫tan2⁡x−cot2⁡xsec2⁡xdx Let tan⁡x=t⇒sec2⁡xdx=dtI=∫t2−1t2dt=t33+1t+C=tan3⁡x3+cot⁡x+C

∫e2log⁡(tan⁡x)−e2log⁡(cot⁡x)cos2⁡xdx=f(x)+C then fπ4=

I=∫elog⁡tan2⁡x−elog⁡cot2⁡xcos2⁡xdxI=∫tan2⁡x−cot2⁡xsec2⁡xdx Let tan⁡x=t⇒sec2⁡xdx=dtI=∫t2−1t2dt=t33+1t+C=tan3⁡x3+cot⁡x+C

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$\int \frac{e^{2 \log (\tan x)} - e^{2 \log (\cot x)}}{\cos^{2} x} d x = f (x) + C then f (\frac{π}{4}) =$

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$\frac{3}{4}$

$\frac{4}{3}$

none

answer is C.

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Detailed Solution

$\begin{array}{l} I = \int \frac{e^{\log (\tan^{2} x)} - e^{\log \cot^{2} x}}{\cos^{2} x} d x \\ I = \int (\tan^{2} x - \cot^{2} x) \sec^{2} x d x \\ Let \tan x = t \Rightarrow \sec^{2} x d x = d t \\ I = \int t^{2} - \frac{1}{t^{2}} d t = \frac{t^{3}}{3} + \frac{1}{t} + C \\ = \frac{\tan^{3} x}{3} + \cot x + C \end{array}$