Calculus II, Chapt. 8
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- ∫ x⿠dx
- (x^(n + 1) / n + 1) + C
- ∫ e^x dx
- e^x + C
- ∫ sin x dx
- -cos x + C
- ∫ sec² x dx
- tan x + C
- ∫ sec x tan x dx
- sec x + C
- ∫ u dv
- uv - ∫ v du
- ∫ tan x dx
- ln |sec x| + C
- ∫ 1 / x dx
- ln |x| + C
- ∫ a ^x dx
- (a^x / ln a) + C
- ∫ cos x dx
- sin x + C
- ∫ csc² x dx
- -cot x + C
- ∫ csc x cot x dx
- -csc x + C
- ∫ cot x dx
- ln |sin x| + C
- ∫ 1 / √(a² - x²) dx
- arcsin (x/a) + C
- ∫ 1 / (x² + a²) dx
- 1/a arctan (x/a) + C
- trig identity sin² =
- 1 - cos² x
- trig identity cos² =
- 1 - sin² x
-
trig half angle identity
sin² = - ½ - ½ cos 2x
-
trig half angle identity
cos² = - ½ + ½ cos 2x
- trig identity sec² x
- tan² + 1
-
trig identity
tan² x - sec² x - 1
- Strategy for solving ∫ sin³(x) cos²(x) where sin is odd
- save one sin factor and use sin²x = 1 - cos²x, then substitute u = cos x
- Strategy for solving ∫ sin²(x) cos³(x) where cos is odd
- save one cos factor and use cos²x = 1 - sin²x, then substitute u = sin x
- Strategy for solving ∫ sin²(x) cos²(x) where both are even
- use half angle identities sin²x = ½(1 - cos 2x) or cos²x = ½(1 + cos 2x)
- Strategy for solving ∫ tan³(x) sec²(x) where tan is odd
- save a factor of sec x tan x and use tan²x = sec²x - 1, then substitute u = sec x
- Strategy for solving ∫ tan³(x) sec²(x) where sec is even
- save a factor of sec²x and use sec²x = 1 + tan²x, then substitute u = tan x
- ∫ csc x dx
- ln |csc x - cot x| + C
- ∫ sec x dx
- ln |sec x + tan x| + C
-
trig identity
sin A cos B - ½[sin(A - B) + sin(A + B)]
-
trig identity
sin A sin B - ½[cos(A - B) - cos(A + B)]
-
trig identity
cos A cos B - ½[cos(A - B) + cos(A + B)
-
trig identity
sin x cos x - ½ sin 2x
- √(a² - x²)
-
x = a sin Θ; dx = a cos Θ dΘ
Θ = sin^-1 x/a
-π/2 - π/2
-> 1 - sin² Θ = cos² Θ - √(a² + x²)
-
x = a tan Θ; dx = a sec² Θ dΘ
Θ = tan ^-1 x/a
-π/2 - π/2
-> 1 + tan² Θ = sec² Θ - √(x² - a²)
-
x = a sec Θ; dx = a sec Θ tan Θ dΘ
0 - π/2
-> sec² Θ - 1 = tan² Θ - sin 2Θ
- 2sin Θ cos Θ
- ∫ sin² x
- ½x - ¼sin(2x)
- ∫ cos² x
- ½x + ¼sin(2x)
- ∫ tan² x
- tan(x) - x