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Test Index
EAMCET Engineering 2008 Solved Paper
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Section:
Mathematics
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© examsnet.com
Question : 154 of 160
Marks:
+1
,
-0
∫
0
1
x
3
∕
2
√
1
−
x
dx
is equal to
π
6
π
9
π
12
π
16
Validate
Solution:
Let
I
=
∫
0
1
x
3
∕
2
√
1
−
x
dx
Put
x
=
s
i
n
2
θ
⇒
dx
=
2
s
i
n
θ
cos
θ
d
θ
∴
I
=
∫
0
π
∕
2
s
i
n
3
θ
⋅
√
1
−
s
i
n
2
θ
2
s
i
n
θ
cos
θ
d
θ
=
2
∫
0
π
∕
2
s
i
n
4
θ
cos
2
θ
d
θ
=
2
[
3
⋅
1
⋅
1
6
⋅
4
⋅
2
⋅
π
2
]
(using Wallis formula)
=
π
16
© examsnet.com
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