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IIT JEE Advanced 2011 Paper 1
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© examsnet.com
Question : 50
Total: 69
The value of
√
l
n
3
∫
√
l
n
2
x
s
i
n
x
2
s
i
n
x
2
+
s
i
n
(
l
n
6
−
x
2
)
d
x
is
1
4
l
n
3
2
1
2
l
n
3
2
l
n
3
2
1
6
l
n
3
2
Validate
Solution:
x
2
= t ⇒ 2x dx = dt
I =
1
2
l
n
3
∫
l
n
2
s
i
n
t
s
i
n
t
+
s
i
n
(
l
n
6
−
t
)
d
t
and =
1
2
l
n
3
∫
l
n
2
s
i
n
(
l
n
6
−
t
)
s
i
n
(
l
n
6
−
t
)
+
s
i
n
t
d
t
2I =
1
2
l
n
3
∫
l
n
2
1
d
t
⇒ I =
1
4
l
n
3
2
.
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