I=∫sec5⁡xtan⁡xdx=∫sec4⁡x(sec⁡xtan⁡x)dxsec⁡x=t→sec⁡xtan⁡xdx=dt=∫t4dt=t55+C=(sec⁡x)55+C

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$\int \sec^{5} x \tan x d x =$

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$\frac{{(\sec x)}^{5}}{5} + C$

$\frac{{(\sec x)}^{6}}{6} + C$

$\frac{{(\sec x)}^{4}}{4} + C$

None

answer is A.

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Detailed Solution

$\begin{array}{l} I = \int \sec^{5} x \tan x d x \\ = \int \sec^{4} x (\sec x \tan x) d x \\ \sec x = t \to \sec x \tan x d x = d t \\ = \int t^{4} d t = \frac{t^{5}}{5} + C \\ = \frac{(\sec x)^{5}}{5} + C \end{array}$

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