Sec x dx = sec x (secx + tanx)/ (secx + tanx) dx . therefore the answer is ln |secx + tanx|
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sec x = 1/cos x so sec x * cos x = 1
The derivative of sec(x) is sec(x) tan(x).
Note that for sec²(x) - tan²(x) = 1, we have: -tan²(x) = 1 - sec²(x) tan²(x) = sec²(x) - 1 Rewrite the expression as: ∫ (sec²(x) - 1) dx = ∫ sec²(x) dx - ∫ 1 dx Finally, integrate each expression to get: tan(x) - x + K where K is the arbitrary constant
Yes, it is. the basic identity is for a double angle relation: cos 2x = 2 cosx cos x -1 since sec x =1/cos x if we multiply both sides by sec x we get cos2xsec x = 2cosxcos x/cos x -1/cos x = 2cos x - sec x
1/cos(x)=sec(x). sec is short for secant.