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Evaluation of definite integrals
Question
∫
0
π
/
2
1
1
+
4
sin
2
x
d
x
=
Easy
A
π
2
5
B
π
5
C
2
π
D
5
π
Solution
∫
0
π
/
2
1
C
o
s
2
x
+
5
S
i
n
2
x
d
x
=
π
2
a
b
=
π
2.1
5
=
π
2
5
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