find int 0 to 2 (4-x²)³/²dx
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Answer:
How do you integrate
∫
(
x
2
√
4
−
x
2
)
d
x
?
Calculus Integrals of Trigonometric Functions
1 Answer
Guillaume L.
Aug 12, 2018
∫
x
2
√
4
−
x
2
d
x
=
2
sin
−
1
(
x
2
)
−
x
2
√
4
−
x
2
+
C
,
C
∈
R
Explanation:
I
=
∫
x
2
√
4
−
x
2
d
x
Let
x
=
2
sin
(
θ
)
d
x
=
2
cos
(
θ
)
d
θ
So:
I
=
∫
4
sin
(
θ
)
2
√
4
−
4
sin
(
θ
)
2
⋅
2
cos
(
θ
)
d
θ
=
2
∫
2
sin
(
θ
)
2
cos
(
θ
)
√
1
−
sin
(
θ
)
2
d
θ
Because
1
−
sin
(
θ
)
2
=
cos
(
θ
)
2
,
I
=
2
∫
2
sin
(
θ
)
2
d
θ
Because
2
sin
(
θ
)
2
=
1
−
cos
(
2
θ
)
I
=
2
∫
(
1
−
cos
(
2
θ
)
)
d
θ
=
2
θ
−
sin
(
2
θ
)
=
2
(
θ
−
sin
(
θ
)
cos
(
θ
)
)
Finally, because
θ
=
sin
−
1
(
x
2
)
, and
cos
(
sin
−
1
(
x
)
)
=
√
1
−
x
2
I
=
2
sin
−
1
(
x
2
)
−
x
√
1
−
x
2
4
=
2
sin
−
1
(
x
2
)
−
x
2
√
4
−
x
2
+
C
,
C
∈
R
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