------------
Welcome to March Calculus Madness!
------------
∫ 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
All original content copyright, © 2011-2022. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.