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$\int (\sec x)^{n} \tan x d x is equal to$

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−(secx)n+1n+1+C

(secx)n+1n+1+C

(secx)nn+C

−1n(secx)n+C

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detailed solution

Correct option is C

Let I=∫(sec⁡x)ntan⁡xdxI=∫(sec⁡x)n−1(sec⁡xtan⁡x)dx Let sec⁡x=tsec⁡xtan⁡xdx=dtI=∫tn−1dt=tnn+C=(sec⁡x)nn+C

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∫(sec⁡x)ntan⁡xdx is equal to

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$\int (\sec x)^{n} \tan x d x is equal to$