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IIT JEE Advanced 2009 Paper 2
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© examsnet.com
Question : 47
Total: 57
If
I
n
=
π
∫
−
π
s
i
n
n
x
(
1
+
π
x
)
s
i
n
x
d
x
, n = 0 , 1 , 2 , ... then
I
n
=
I
n
+
2
10
Σ
m
=
1
I
m
+
1
= 10π
10
Σ
m
=
1
I
2
m
= 0
I
n
=
I
n
+
1
Validate
Solution:
(A) We have the integral
I
n
=
π
∫
−
π
s
i
n
n
x
(
1
+
π
x
)
s
i
n
x
d
x
=
π
∫
0
(
s
i
n
n
x
(
1
+
π
x
)
s
i
n
x
+
π
x
s
i
n
n
x
(
1
+
π
x
)
s
i
n
x
)
d
x
=
π
∫
0
s
i
n
n
x
s
i
n
x
Now,
I
n
+
2
−
I
n
=
π
∫
0
s
i
n
(
n
+
2
)
x
−
s
i
n
n
x
s
i
n
x
d
x
(B) Since
I
3
=
I
5
= ...
I
21
, we have
10
Σ
m
=
1
I
2
m
+
1
=
10
I
3
=
10
π
∫
0
s
i
n
3
x
s
i
n
x
d
x
=
10
π
∫
0
(
3
−
4
s
i
n
2
x
)
d
x
=
10
[
3
x
−
2
x
+
s
i
n
2
x
]
0
π
= 2π
(C) Since
I
2
=
I
4
= .. =
I
20
, we have
10
Σ
m
=
1
I
2
m
=
10
π
∫
0
s
i
n
2
x
s
i
n
x
d
x
=
20
[
s
i
n
x
]
0
π
= 0
© examsnet.com
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