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JEE Main Math Class 12 Integrals Part 4 Questions
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© examsnet.com
Question : 4
Total: 51
The value of
π
∫
−
π
2
y
(
1
+
s
i
n
y
)
1
+
cos
2
y
d
y
is:
[5 Apr 2024 Shift 1]
π
2
π
2
2
π
2
2
π
2
Validate
Solution:
👈: Video Solution
π
∫
−
π
2
y
(
1
+
s
i
n
y
)
1
+
cos
2
y
d
y
=
π
∫
−
π
2
y
1
+
cos
2
y
d
y
+
π
∫
−
π
2
y
s
i
n
y
1
+
cos
2
y
d
y
=
0
+
2.2
π
∫
0
y
(
s
i
n
y
1
+
cos
2
y
)
d
y
I
=
4
π
∫
0
y
s
i
n
y
1
+
cos
2
y
d
y
I
=
4
π
∫
0
(
π
−
y
)
s
i
n
y
1
+
cos
2
y
d
y
2
I
=
4
π
∫
0
π
s
i
n
y
1
+
cos
2
y
d
y
I
=
2
π
π
∫
0
s
i
n
y
1
+
cos
2
y
d
y
=
2
π
(
−
tan
−
1
(
cos
y
)
)
0
π
=
−
2
π
[
(
−
π
4
)
−
(
π
4
)
]
=
−
2
π
[
−
2
π
4
]
=
π
2
© examsnet.com
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