X by the substitution of du, d u, and be left with an even number of cosine powers Use cos2x = 1−sin2x (= 1−u2) cos 2 x = 1 − sin 2 x ( = 1 − u 2) to replace the leftover cosines Both m m and n n are odd Use either 1 1 or 2 2 (both will work) Both m m and n n are even Use cos2x = 1 2(1cos(2x)) cos 2 Sec(x) tan(x) dx = Z 1 sec(x) tan(x) sec2(x) sec(x)tan(x) dx = Z 1 u du for ˆ u= sec( x) tan( ) du= (sec2( x) sec( )tan( ))dx = lnjuj C = lnjsec(x)tan(x)j C Another trick for this is to write R sec(x)dx= R 1 cos2(x) cos(x)dx, and substitute u= sin(x) to get R 1 1 u2 du We will see how to integrate such rational functions in x74 $\begingroup$ I actually prefer Using $\tan$ and $\sec$ because they seem much simpler to me than converting to rational functions $(\tan x)' = \sec^2 x$ and $(\sec x)' = \sec x \tan x$ go handinhand $\endgroup$ – Dylan Sep 16 '15 at 21
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