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UPSC NDA Math Model Paper 2
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© examsnet.com
Question : 14
Total: 120
π
∫
0
x
f
(
sin
x
)
d
x
is equal to:
π
π
∕
2
∫
0
f
(
sin
x
)
d
x
π
/
2
π
∕
2
∫
0
f
(
sin
x
)
d
x
π
π
∕
2
∫
0
f
(
cos
x
)
d
x
π
2
π
∫
0
f
(
cos
x
)
d
x
Validate
Solution:
Let
I
=
π
∫
0
x
f
(
sin
x
)
d
x
Using
b
∫
a
f
(
x
)
d
x
=
b
∫
a
f
(
a
+
b
−
x
)
d
x
,
we get:
⇒
I
=
π
∫
0
(
π
−
x
)
f
[
sin
(
π
−
x
)
]
d
x
⇒
I
=
π
∫
0
π
f
(
sin
x
)
d
x
−
π
∫
0
x
f
(
sin
x
)
d
x
⇒
2
I
=
π
π
∫
0
f
(
sin
x
)
d
x
Since
f
[
sin
(
π
−
x
)
]
=
f
(
sin
x
)
,
using
2
a
∫
0
f
(
x
)
d
x
=
2
a
∫
0
f
(
x
)
d
x
,
we get
⇒
2
I
=
2
π
π
∕
2
∫
0
f
(
sin
x
)
d
x
⇒
I
=
π
π
∕
2
∫
0
f
(
sin
x
)
d
x
© examsnet.com
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