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Mathematics
Mathematics
Problem
If
∫
0
π
(
x
1
+
sin
x
)
2
d
x
=
A
\int_0^\pi \left(\frac{x}{1+\sin x}\right)^2 dx = A
∫
0
π
(
1
+
s
i
n
x
x
)
2
d
x
=
A
, then the value for
∫
0
π
2
x
2
⋅
cos
2
x
/
2
(
1
+
sin
x
)
2
d
x
\int_0^\pi \frac{2x^2 \cdot \cos^2 x/2}{(1+\sin x)^2} dx
∫
0
π
(
1
+
s
i
n
x
)
2
2
x
2
⋅
c
o
s
2
x
/2
d
x
is equal to a.
A
+
2
π
−
π
2
A + 2\pi - \pi^2
A
+
2
π
−
π
2
b.
A
−
2
π
+
π
2
A - 2\pi + \pi^2
A
−
2
π
+
π
2
c.
2
π
−
A
−
π
2
2\pi - A - \pi^2
2
π
−
A
−
π
2
d. None of these
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