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JEE Main 9 Apr 2024 Shift 2 Solved Paper
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© examsnet.com
Question : 15
Total: 90
The integral
3
∕
4
∫
1
∕
4
cos
(
2
cot
−
1
√
1
−
x
1
+
x
)
dx
is equal to:
[9 Apr 2024 Shift 2]
−
1
∕
2
1
∕
4
1
∕
2
−
1
∕
4
Validate
Solution:
👈: Video Solution
I
=
3
∕
4
∫
1
∕
4
cos
(
2
cot
−
1
(
√
1
−
x
1
+
x
)
dx
)
3
∕
4
∫
1
∕
4
cos
(
2
(
tan
−
1
√
1
+
x
1
+
x
)
)
dx
3
∕
4
∫
1
∕
4
1
−
tan
2
(
tan
−
1
√
1
+
x
1
−
x
)
tan
2
(
tan
−
1
√
1
+
x
1
−
x
)
dx
=
3
∕
4
∫
1
∕
4
1
−
(
1
+
x
1
−
x
)
1
+
(
1
+
x
1
−
x
)
dx
=
3
∕
4
∫
1
∕
4
−
2
x
2
dx
=
3
∕
4
∫
1
∕
4
(
−
x
)
dx
=
−
(
x
2
2
)
1
∕
4
3
∕
4
=
−
1
2
[
9
16
−
1
16
]
=
−
1
4
© examsnet.com
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