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Test Index
JEE Main 1-Feb-2024 Shift 1 Solved Paper
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Section:
Mathematics
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© examsnet.com
Question : 2
Total: 90
The value of the integral
π
4
∫
0
x
dx
s
i
n
4
(
2
x
)
+
cos
4
(
2
x
)
equals :
[1-Feb-2024 Shift 1]
√
2
π
2
8
√
2
π
2
16
√
2
π
2
32
√
2
π
2
64
Validate
Solution:
👈: Video Solution
π
4
∫
0
x
dx
s
i
n
4
(
2
x
)
+
cos
4
(
2
x
)
Let
2
x
=
t
then
d
x
=
1
2
dt
I
=
1
4
π
2
∫
0
t
dt
s
i
n
4
t
+
cos
4
t
I
=
1
4
π
2
∫
0
(
π
2
−
t
)
dt
s
i
n
4
(
π
2
−
t
)
+
cos
4
(
π
2
−
t
)
I
=
1
4
π
2
∫
0
π
2
dt
s
i
n
4
t
+
cos
4
t
−
I
2
I
=
π
8
π
2
∫
0
d
t
s
i
n
4
t
+
cos
4
t
2
I
=
π
8
π
2
∫
0
s
e
c
4
t
dt
tan
4
t
+
1
Let
tan
=
y
then
sec
2
tdt
=
dy
2
I
=
π
8
∞
∫
0
(
1
+
y
2
)
d
y
1
+
y
4
=
π
16
∞
∫
0
1
+
1
y
2
y
2
+
1
y
2
dy
Put
y
−
1
y
=
p
I
=
π
16
∞
∫
−
∞
dp
p
2
+
(
√
2
)
2
=
π
16
√
2
[
tan
−
1
(
p
√
2
)
]
−
∞
∞
I
=
π
2
16
√
2
© examsnet.com
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