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KEAM 2018 Math Paper
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© examsnet.com
Question : 97
Total: 120
π
2
∫
0
2
s
i
n
x
2
s
i
n
x
+
2
c
o
s
x
dx
=
2
π
π
4
2
π
0
Validate
Solution:
Let
I
=
∫
0
π
2
2
sin
x
2
sin
x
+
2
cos
x
d
x
.(i)
⇒
I
=
∫
2
π
2
2
sin
(
π
2
−
x
)
2
sin
(
π
2
−
x
)
+
2
cos
(
π
2
−
x
)
⇒
I
=
∫
0
π
2
2
cos
x
2
cos
x
+
2
sin
x
....(ii)
By adding Eqs. (i) and (ii), we get
2
I
=
∫
0
π
2
2
sin
x
+
2
cos
x
2
sin
x
+
2
cos
x
d
x
⇒
2
I
=
∫
0
π
2
1
⋅
d
x
=
[
x
]
0
π
2
⇒
2
I
=
π
2
⇒
I
=
π
4
© examsnet.com
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