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AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper
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© examsnet.com
Question : 77
Total: 160
∞
∫
0
e
−
x
s
i
n
6
x
dx =
24
85
124
285
136
529
144
629
Validate
Solution:
The formula for the integration,
I
m
=
∞
∫
0
e
−
x
sin
m
x
d
x
,
where
m
>
2
then
I
m
=
m
(
m
−
1
)
1
+
m
2
I
m
−
2
N
o
w
I
6
=
∞
∫
0
e
−
x
sin
6
x
d
x
I
6
=
6
(
5
)
1
+
6
2
I
4
I
6
=
30
37
I
4
Similarly,
I
4
=
∞
∫
0
e
−
x
sin
4
x
d
x
I
4
=
4
(
3
)
1
+
4
2
I
2
I
4
=
12
17
I
2
Similarly
I
6
=
∞
∫
0
e
−
x
sin
6
x
d
x
I
2
=
2
(
1
)
1
+
2
2
I
0
I
2
=
2
5
I
0
And
I
0
=
∞
∫
0
e
−
x
sin
0
x
d
x
I
0
=
[
−
e
−
x
]
0
∞
=
1
The value of
I
6
is,
I
6
=
30
37
×
12
17
×
2
5
×
1
=
144
629
© examsnet.com
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