Inverse trig functions are best integrated by parts. The following
trig identities are also important:
sin² x + cos² x = 1.
sec² x = tan² x + 1.
cosec² x = cot² x + 1.
1. ∫ Arc sin x dx.
Let u= arc sin x and dv = dx.
Then du= 1/(√1-x² ) dx and v=x.
∫ arc sin x dx = ∫ u dv = uv - ∫ v du = x arc sin x - ∫ x/(√1-x² )
dx= x arc sin x + √ (1-x²) + C.
NOTE: Since u= arc sin x implies x= sin u, ∫ x/(√1-x² ) dx = ∫(sin u)
/ √ (1- sin² u) . Cos u du = ∫[(sin u)/(cos u)]. Cos u du = ∫ sin u du
= - cos u = - √ (1- sin² u) = - √ (1 - x²).
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2. ∫ arc cos x dx.
Let u = arc cos x and dv = dx.
Then du = -1 /√ (1- x²) dx and v = x.
Hence, ∫ arc cos x dx = ∫ u dv = uv - ∫ v du = x arc cos x - ∫ -x/√
(1-x²) dx = x arc cos x + ∫x/√(1-x²) dx = x arc cos x - √ (1-x²)
+ C.
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3. ∫ Arc tan x dx.
Let u = arc tan x and dv = dx.
Then du= 1/(x² + 1) dx and v = x.
Therefore, ∫ arc tan x dx = ∫u dv = uv -∫ v du = x arc tan x - ∫ x/(x²
+ 1) dx = x arc tan x - (1/2) In (x² + 1) + C.
NOTE: ∫ x/(x² + 1) dx = ∫[(1/2) x 2x]/ (x² + 1) dx = (1/2) ∫ (2x)/(x²
+ 1) = (1/2) In (x² + 1) , since ∫f'(x)/f(x) dx = In f(x).
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4. Arc sec x.
Let u= arc sec x and dv = dx.
Then sec u = x ; dx/du = tan u sec u = sec u X √ (sec² u - 1) = x√(x²
- 1) ; du/dx = 1 /(dx/du) = 1/[x√(x² - 1) ] ; du = 1/[x√(x² - 1) ]
dx. V=
∫ dx = x.
Therefore, ∫ arc sec x dx = x arc sec x - ∫x . 1/[x√(x² - 1) ] dx =
x arc sec x - ∫ 1/√(x² - 1) dx = x arc sec x - In [x + √(x² - 1)] +
C.
NOTE: Since x = sec u, dx/du = tan u sec u ; dx= tan u sec u du.
Substituting, we have:
∫1/√ (x² - 1) dx = ∫ sec u du = ∫ sec u X [ (tan u + sec u)/ (tan u +
sec u)] du = ∫ (sec u tan u + sec² u)/ (tan u + sec u) du = In (tan u
+ sec u) = In [(x + √ (x² - 1)].
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5. ∫ arc cosec x dx.
Let u = arc cosec x and dv =dx.
Then du= -1 /[x√(x²-1)] dx and v = x.
Therefore, ∫ arc cosec x dx = x arc cosec x - ∫ x. -1 [x√(x²-1)] dx =
x arc cosec x + ∫ 1/√ (x²-1) dx = x arc cosec x + In [x + √(x²
- 1)] + C.
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6. ∫ arc cot x dx.
Let u = arc cot x and dv = dx.
Then, du = -1/ (x² + 1) dx and v= x.
Therefore, ∫ arc cot x dx = x arc cot x - ∫ x . -1/(x² + 1) dx = x
arc cot x + ∫ x/(x² + 1) dx = x arc cot x + (1/2) In (x² + 1) + C.
(See the workings for the integral @ arc tan x above).
NOTE: u = arc cot x implies x = cot u = cos u/sin u. Then, using
Quotient rule, dx/du = -1 - cot² u = - 1 - x². Hence, du/dx = 1 / (-1
-x²) = -1 / (x² + 1);...; du = -1/(x² + 1) dx.
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