integrate[sec^3x]

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integrate[sec^3x]

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[\int\limits_{?}^{?}\sec ^{3}\theta d \theta\]
|dw:1327731874453:dw||dw:1327732061041:dw|
because the reciprocal ratio for sec is 1/cosx

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so then just bring the power to the front
so
thanks, never thought about that move;
|dw:1327732192064:dw|
also have to know the general power rule wich is
|dw:1327732252138:dw|
im not getting the rule.. explain?
no that won't quite work above you are going to integration by parts I think here
it was solved, but idk y he squared the last integral-- and the power rule he stated
u=secx du=secxtanx dv=sec^2x u=tanx
\[\int \sec^3xdx=\sec x \tan x-\int \sec x\tan^2xdx\]\[=\sec x \tan x-\int \sec x(1-\sec^2x)dx\]\[=\sec x \tan x-\int\sec^3 xdx+\int\sec xdx\]have it from there ?
v=tanx above*
yes, got it; thanks, for the integral of secx, you change to 1/cosx, then multiply by cosx/cos right?
\[\int\sec xdx=\int\sec x(\frac{\sec x+\tan x}{\sec x+\tan x})=\int\frac{du}{u}\]
i c, kk thanks alot man =)
anytime

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