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Manipal Entrance Test 2017 Math Paper
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© examsnet.com
Question : 4
Total: 70
1
∫
0
d
d
x
[
sin
−
1
(
2
x
1
+
x
2
)
]
d
x
is equal to
0
π
π
2
π
4
Validate
Solution:
Let
I
=
1
∫
0
d
d
x
[
sin
−
1
(
2
x
1
+
x
2
)
]
d
x
∴
I
=
1
∫
0
d
d
x
[
sin
−
1
(
sin
2
θ
)
]
d
x
(put x =
tan
θ
)
=
2
1
∫
0
d
d
x
[
tan
−
1
x
]
d
x
=
2
[
tan
−
1
(
1
)
−
tan
−
1
(
0
)
]
=
2
.
π
4
=
π
2
© examsnet.com
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