Evaluate $ \int \cfrac { \sec^2x}{ ( \sec x + \tan x)^{9/2}}dx $

Asked Sep 06 '13 at 05:17

Active Sep 07 '13 at 05:25

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Problem : Evaluate $ \int \cfrac { \sec^2x}{ ( \sec x + \tan x)^{9/2}}dx $

Solution : $ \int \cfrac { \sec^2x}{ ( \sec x + \tan x)^{9/2}}dx $

$ \int \cfrac { \sec^2x}{ \left(1+ \sin x \over \cos x \right)^ {9/2}}dx $

$ \int \cfrac { \sec^2x}{ (\sin {x \over 2} + \cos {x \over 2})^9 \over (\cos^2 {x \over 2} - \sin^2 {x \over 2})^ {9/2}}dx $

$ \int \cfrac { \sec^2x}{ \left(\cos {x \over 2} + \sin {x \over 2} \over\cos {x \over 2} - \sin {x \over 2}\right)^ {9/2}}dx $

$ \int \cfrac { \sec^2x}{ \left(1 + \tan {x \over 2} \over 1 - \tan {x \over 2}\right)^ {9/2}}dx $

$ \int \cfrac { \sec^2x}{[tan({ \pi \over 4}+ { x \over 2})]^ {9/2}}dx $

Am I doing right ?

edited Sep 07 '13 at 05:25

asked Sep 06 '13 at 05:17

rst

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1

See http://math.stackexchange.com/questions/130987/evaluate-int-frac-sec2x-secx-tanx5-2-mathrm-dx?rq=1 – G Tony Jacobs Sep 06 '13 at 05:27
thanks.Above link is useful – rst Sep 06 '13 at 05:39

Evaluate $ \int \cfrac { \sec^2x}{ ( \sec x + \tan x)^{9/2}}dx $

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