∫ 0 1 / 2 ( 1 − x 2 ) 3 / 2 x 2 d x
Given that the integral above is of the form : C A B − π , where A and C are positive integers, B is a square-free integer, find A + B + C .
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i got the same thing,but i got confused because it said that A and C are coprime, 2 and 6 are obviously not coprime
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P
=
∫
0
2
1
2
(
1
−
x
2
)
2
3
x
⋅
2
x
d
x
Integrating by parts,
P
=
⎣
⎢
⎡
2
1
−
x
2
2
1
x
⎦
⎥
⎤
0
2
1
−
∫
0
2
1
(
d
x
d
x
)
2
1
−
x
2
2
1
1
d
x
∴
P
=
3
1
−
[
sin
−
1
x
]
0
2
1
∴
P
=
3
1
−
6
π
=
6
2
3
−
π
Comparing,
A
=
2
,
B
=
3
,
C
=
6
A
+
B
+
C
=
2
+
3
+
6
=
1
1
Great explanation.
What inspired the choice of variables for the integration by parts?
I noticed that if I write x 2 = x ⋅ x , and integrate by parts, one of them gets eliminated by differentiation and the other allows to integrate the denominator.
...
The indefinite integral is ,
y − a r c t a n ( y ) + C = t a n ( t ) − t + C = 1 − s i n 2 ( t ) s i n ( t ) − t + C = 1 − x 2 x − a r c s i n ( x ) ) + C for x ϵ ( − 1 , 1 )
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I got the same result I just put in the limits in different steps.
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Using the trigonometric substitution x = sin θ , d x = cos θ d θ we transform the integral like so:
∫ ( 1 − x 2 ) 2 3 x 2 d x ⟶ ∫ cos 3 θ sin 2 θ cos θ d θ = ∫ tan 2 θ d θ
Using the pythagorean identity 1 + tan 2 θ = sec 2 θ , we rewrite the resulting integral and integrate:
∫ tan 2 θ d θ = ∫ sec 2 θ − 1 d θ = tan θ − θ
Convert this result back to the variable x to obtain
1 − x 2 x − arcsin x ∣ ∣ ∣ 0 2 1 = 3 3 − 6 π = 6 2 3 − π
Here, A = 2, B = 3, C = 6, so our answer is 2 + 3 + 6 = 1 1 .