integral of xsinxcosx/(a^2 cos^2x + b^2 sin^2x)^2 from (0 to pi/2)
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Step-by-step explanation:
Correct option is
A
4
Let I=∫
0
π
(a
2
cos
2
x+b
2
sin
2
x)
2
xdx
Applying integration property f(a−x)=f(x)
2I=∫
0
π
(a
2
cos
2
+x+b
2
sin
2
x)
2
πdx
=2π∫
0
π/2
(a
2
cos
2
x+b
2
sin
2
x)
2
dx
Using integration property f(2a−x)=f(x)
Therefore
I=π∫
0
π/2
(a
2
+b
2
tan
2
x)
2
sec
2
xsec
2
xdx
Put btanx=atanθ⇒bsec
2
xdx=asec
2
θdθ
⇒sec
2
xdx=
b
a
(1+tan
2
θ)dθ
Therefore
I=π∫
0
π/2
a
4
(1+tan
2
θ)
2
(1+tan
2
x)
b
a
(1+tan
2
θ)dθ
=
a
3
b
π
∫
0
π/2
(1+
b
2
a
2
tan
2
θ)cos
2
θdθ
=
a
3
b
3
π
∫
0
π/2
(b
2
cos
2
θ+a
2
sin
2
θ)dθ
=
a
3
b
3
π
[
2
1
,
2
π
](b
2
+a
2
)=
4
π
2
a
3
b
3
a
2
+b
2
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