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JEE Main 6 Apr 2024 Shift 2 Solved Paper
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© examsnet.com
Question : 19
Total: 90
If
∫
1
a
2
s
i
n
2
x
+
b
2
cos
2
x
dx
=
1
12
tan
−
1
(
3
tan
x
)
+
constant, then the maximum value of asin
x
+
b
cos
x
, is :
[6 Apr 2024 Shift 2]
√
40
√
39
√
42
√
41
Validate
Solution:
👈: Video Solution
∫
s
e
c
2
x
dx
a
2
tan
2
x
+
b
2
let
tan
x
=
t
s
e
c
2
dx
=
d
t
∫
d
t
a
2
t
2
+
b
2
1
a
2
∫
d
t
t
2
+
(
b
a
)
2
1
a
2
1
b
a
tan
−
1
(
t
b
a
)
+
c
1
a
b
tan
−
1
(
α
b
tan
x
)
+
c
on comparing
a
b
=
3
a
b
=
12
a
=
6
,
b
=
2
maximum value of
6
s
i
n
x
+
2
cos
x
is
√
40
© examsnet.com
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