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JEE Main Math Class 12 Integrals Part 2 Questions
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© examsnet.com
Question : 4
Total: 100
The value of the definite integral
π
4
∫
−
π
4
d
x
(
1
+
e
x
cos
x
)
(
sin
4
x
+
cos
4
x
)
is equal to:
[27 Jul 2021 Shift 1]
−
π
2
π
2
√
2
−
π
4
π
√
2
Validate
Solution:
π
4
∫
−
π
4
d
x
(
1
+
e
x
cos
x
)
(
sin
4
x
+
cos
4
x
)
.........(1)
Using
b
∫
a
f
(
x
)
d
x
=
b
∫
a
f
(
a
+
b
−
x
)
d
x
I
=
π
4
∫
−
π
4
d
x
(
1
+
e
−
x
cos
x
)
(
sin
4
x
+
cos
4
x
)
........(2)
Add (1) and (2)
2I
=
π
4
∫
−
π
4
d
x
sin
4
x
+
cos
4
x
2
I
=
2
π
4
∫
0
d
x
sin
4
x
+
cos
4
x
I
=
π
4
∫
0
(
1
+
tan
2
x
)
sec
2
x
tan
4
x
+
1
d
x
I
=
π
4
∫
0
(
1
+
1
tan
2
x
)
sec
2
x
(
tan
x
−
1
tan
x
)
2
+
2
d
x
tan
x
−
1
tan
x
=
t
(
1
+
1
tan
2
x
)
sec
2
x
d
x
=
d
t
I
=
0
∫
−
∞
d
t
t
2
+
2
=
[
1
√
2
tan
−
1
(
t
√
2
)
]
−
∞
0
I
=
0
−
1
√
2
(
−
π
2
)
=
π
2
√
2
© examsnet.com
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