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Evaluation of definite integrals
Question
∫
0
π
sin
θ
+
cos
θ
1
+
sin
2
θ
d
θ
=
Easy
A
π
B
π
+
λ
and
λ
>
0
C
π
/
2
D
π
/
3
Solution
∫
0
π
sin
θ
+
cos
θ
1
+
sin
2
θ
d
θ
=
∫
0
π
sin
θ
+
cos
θ
|
sin
θ
+
cos
θ
|
d
θ
=
π
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