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IIT JEE Advanced 2008 Paper 2
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Section:
Mathematics
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© examsnet.com
Question : 46
Total: 66
The area of the region between the curves
y =
√
1
+
s
i
n
x
c
o
s
x
and y =
√
1
−
s
i
n
x
c
o
s
x
bounded by the lines x = 0 and x =
Ï€
4
is
[JEE Adv 2008 P2]
√
2
−
1
∫
0
t
(
1
+
t
2
)
√
1
−
t
2
√
2
−
1
∫
0
4
t
(
1
+
t
2
)
√
1
−
t
2
√
2
+
1
∫
0
4
t
(
1
+
t
2
)
√
1
−
t
2
√
2
+
1
∫
0
t
(
1
+
t
2
)
√
1
−
t
2
Validate
Solution:
Since both curves lie above x-axis in x ∊
(
0
,
Ï€
4
)
. The area bounded between the curves is
=
Ï€
4
∫
(
√
1
+
s
i
n
x
c
o
s
x
−
√
1
−
s
i
n
x
c
o
s
x
)
d
x
=
Ï€
4
∫
0
(
√
1
+
t
a
n
x
2
1
−
t
a
n
x
2
−
√
1
−
t
a
n
x
2
1
+
t
a
n
x
2
)
d
x
=
Ï€
4
∫
0
(
2
t
a
n
x
2
√
1
−
t
a
n
2
x
2
)
d
x
Substituting tan
x
2
= t , we get
1
2
s
e
c
2
x
2
d
x
= dt
dx =
2
d
t
1
+
t
2
dx =
2
d
t
1
+
t
2
=
√
2
−
1
∫
0
(
4
t
(
1
+
t
2
)
√
1
−
t
2
)
d
t
© examsnet.com
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