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Question
Mathematics
∫ limits1e(17/2) ( π cos (π log x)/x) dx =
Q.
1
∫
e
2
17
x
π
c
o
s
(
π
l
o
g
x
)
d
x
=
2228
236
COMEDK
COMEDK 2010
Integrals
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A
0
33%
B
-1
25%
C
2
25%
D
1
16%
Solution:
Let
I
=
1
∫
e
2
17
x
π
c
o
s
(
π
l
o
g
x
)
d
x
Putting
p
i
lo
g
x
=
z
⇒
x
π
d
x
=
d
z
Since,
1
≤
x
≤
e
17/2
⇒
0
≤
z
≤
2
17
π
⇒
I
=
∫
0
17
π
/2
cos
z
d
x
=
[
sin
z
]
0
17
π
/2
=
sin
2
17
π
−
sin
0
=
sin
(
8
π
+
2
π
)
=
sin
2
π
=
1