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Welcome to March Calculus Madness!
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∫ 1/(1 + cos x) dx and ∫ 1/(1 + sin x) dx
To tackle these integrals, we will make use of the following derivatives and trig identities:
d/dx tan x = sec^2 x dx
d/dx cot x = -csc^2 x dx
d/dx csc x = -cot x * csc x
d/dx sec x = tan x * sec x
sin^2 x + cos^2 x = 1
cot x = 1/tan x, csc x = 1/sin x, sec x = 1/cos x
∫ 1/(1 + cos x) dx
= ∫ 1/(1 + cos x) * (1 - cos x)/(1 - cos x) dx
= ∫ (1 - cos x)/(1 - cos^2 x) dx
= ∫ (1 - cos x)/sin^2 x dx
= ∫ csc^2 x - cot x * csc x dx
= -cot x + csc x + C
∫ 1/(1 + sin x) dx
= ∫ 1/(1 + sin x) * (1 - sin x)/(1 - sin^2 x) dx
= ∫ (1 - sin x)/(1 - sin^2 x) dx
= ∫ (1 - sin x)/(cos^2 x) dx
= ∫ sec^2 x - tan x * sec x dx
= tan x - sec x + C
Note: CAS systems in graphing calculators will default to using sin x, cos x, and tan x.
Eddie
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