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UPSC NDA Math Model Paper 9
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© examsnet.com
Question : 3 of 120
Marks:
+1
,
-0
What is
π
2
∫
0
d
x
a
2
cos
2
x
+
b
2
sin
2
x
equal to?
2ab
2πab
π
2
a
b
π
a
b
Validate
Solution:
Let,
I
=
π
2
∫
0
d
x
a
2
cos
2
x
+
b
2
sin
2
x
I
=
π
2
∫
0
d
x
cos
2
x
(
a
2
+
b
2
sin
2
x
cos
2
x
)
=
π
2
∫
0
sec
2
x
d
x
a
2
+
b
2
tan
2
x
Let
tan
x
=
t
⇒
sec
2
x
d
x
=
d
t
When x = 0, t = tan (0) = 0
When x = π/2, t = ∞
I
=
∞
∫
0
d
t
a
2
+
b
2
t
2
=
1
b
2
∞
∫
0
d
t
(
(
a
b
)
2
+
t
2
)
=
1
b
2
[
b
a
tan
−
1
(
b
t
a
)
]
0
∞
=
1
a
b
[
tan
−
1
(
∞
)
−
tan
−
1
(
0
)
]
=
1
a
b
×
(
π
2
−
0
)
=
π
2
a
b
Hence, option (3) is correct.
© examsnet.com
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