Question

sec
${sec}^{5}x \: tanx \: dx$
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Answer 1

Correction in question:

$\\\tt \int\limits sec^{ 5 }x \ tanx \ dx$

Answer:

$\\\tt \dfrac{sec^{5} x}{5} + C$

Step-by-step explanation:

$\\\tt \int\limits sec^{5}x \ tanx \ dx \\\\\tt \rightarrow \int\limits( sec^{4}x)\ (sec x\ tanx) \ dx$    

Let y = sec x

Now,

$\\\tt dy = secx\ tanx\ dx$

Using these values in the integrand,

$\\\tt \int\limits y^{4} ( \dfrac{sec x\ tanx}{sec x \ tan x}) \ dy \\\\ \\= \int\limits y^{4} \ dy \\\\\\= \dfrac{y^{5}}{5} + C$

Inserting the value of y in above result,

$\\\tt \dfrac{sec^{5} x}{5} + C$