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JEE Main 24 June 2022 Shift 2 Solved Paper
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© examsnet.com
Question : 6
Total: 90
The value of the integral
π
∕
2
∫
−
π
∕
2
d
x
(
1
+
e
x
)
(
sin
6
x
+
cos
6
x
)
is equal to
[24-Jun-2022-Shift-2]
2
π
0
π
π
2
Validate
Solution:
👈: Video Solution
I
=
π
2
∫
−
π
2
d
x
(
1
+
e
x
)
(
sin
6
x
+
cos
6
x
)
.
.
.
.
.
.
(i)
I
=
π
2
∫
−
π
2
d
x
(
1
+
e
−
x
)
(
sin
6
x
+
cos
6
x
)
.
.
.
.
.
(ii)
(i) and (ii)
From equation (i) \& (ii)
2
I
=
π
2
∫
−
π
2
d
x
sin
6
x
+
cos
6
x
⇒
I
=
∫
0
π
2
d
x
sin
6
x
+
cos
6
x
=
∫
0
π
2
d
x
1
−
3
4
sin
2
2
x
⇒
I
=
∫
0
π
2
4
sec
2
2
x
d
x
4
+
tan
2
2
x
=
2
∫
0
π
4
4
sec
2
2
x
4
+
tan
2
2
x
d
x
when
x
=
0
,
t
=
0
Now,
tan
2
x
=
t
when
x
=
π
4
,
t
→
∞
2
sec
2
2
x
d
x
=
d
t
∴
I
=
2
∫
0
∞
2
d
t
4
+
t
2
=
2
(
tan
−
1
t
2
)
0
∞
=
2
π
2
=
π
© examsnet.com
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