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Manipal Entrance Test 2020 Math Paper
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© examsnet.com
Question : 12
Total: 70
If
∫
sin
{
2
tan
−
1
√
1
−
x
1
+
x
}
d
x
=
A
sin
−
1
x
+
B
x
√
1
−
x
2
+
C
then
A
+
B
is equal to
10
1
2
1
−
1
2
Validate
Solution:
We have,
∫
sin
{
2
tan
−
1
√
1
−
x
/
1
+
x
}
d
x
=
A
sin
−
1
x
+
B
x
√
1
−
x
2
+
C
Let
I
=
∫
sin
{
2
tan
−
1
√
1
−
x
/
1
+
x
}
d
x
Put
x
=
cos
2
θ
⇒
d
x
=
−
2
sin
2
θ
d
θ
∴
I
=
−
∫
sin
{
2
tan
−
1
√
1
−
cos
2
θ
/
1
+
cos
2
θ
}
2
sin
2
θ
d
θ
=
−
∫
sin
{
2
tan
−
1
√
2
sin
2
θ
/
2
cos
2
θ
}
2
sin
2
θ
d
θ
=
−
∫
sin
{
2
tan
−
1
√
tan
2
θ
}
2
sin
2
θ
d
θ
=
−
∫
sin
{
2
tan
−
1
tan
θ
}
2
sin
2
θ
d
θ
=
−
∫
2
sin
2
2
θ
d
θ
=
−
∫
(
1
−
cos
4
θ
)
d
θ
=
−
[
θ
−
sin
4
θ
4
]
+
C
=
[
−
1
2
cos
−
1
x
+
1
4
×
2
sin
2
θ
cos
2
θ
]
+
C
But
I
=
A
sin
−
1
x
+
B
x
√
1
−
x
2
+
C
∴
A
=
1
2
and
B
=
1
2
Hence,
A
+
B
=
1
2
+
1
2
=
1
© examsnet.com
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