find the value of integration of root of (9-6x ) dx.
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Answer:
Explanation:
sin
2
A
+
cos
2
A
=
1
cos
2
A
=
2
cos
2
A
−
1
sin
2
A
=
2
sin
A
cos
A
Use substitution.
x
=
3
sin
t
∫
√
9
−
x
2
d
x
=
∫
3
√
1
−
sin
2
t
×
3
cos
t
d
t
=
9
∫
cos
2
t
d
t
9
∫
cos
2
t
d
x
=
9
2
∫
(
1
+
cos
2
t
)
d
t
=
9
2
(
t
+
1
2
sin
2
t
)
+
C
In terms of
x
:
∫
√
9
−
x
2
d
x
=
9
2
(
sin
−
1
(
x
3
)
+
x
3
√
1
−
(
x
3
)
2
)
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