find the integration of logxsinx
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Step-by-step explanation:
Answer:
I
=
∫
π
2
0
log
sin
x
d
x
=
−
(
π
2
)
log
2
Explanation:
We use the property
∫
a
0
f
(
x
)
d
x
=
∫
a
0
f
(
a
−
x
)
d
x
hence we can write
I
=
∫
π
2
0
log
sin
x
d
x
=
∫
π
2
0
log
sin
(
π
2
−
x
)
d
x
or
I
=
∫
π
2
0
log
sin
x
d
x
=
∫
π
2
0
log
cos
x
d
x
or
2
I
=
∫
π
2
0
(
log
sin
x
+
log
cos
x
)
d
x
=
∫
π
2
0
log
(
sin
x
cos
x
)
d
x
=
∫
π
2
0
log
(
sin
2
x
2
)
d
x
=
∫
π
2
0
(
log
sin
2
x
−
log
2
)
d
x
=
∫
π
2
0
log
sin
2
x
d
x
−
∫
π
2
0
log
2
d
x
=
∫
π
2
0
log
sin
2
x
d
x
−
(
π
2
)
log
2
.............(A)
Let
I
1
=
∫
π
2
0
log
sin
2
x
d
x
and
t
=
2
x
, then
I
1
=
1
2
∫
π
0
log
sin
t
d
t
and using the property
∫
2
a
0
f
(
x
)
d
x
=
2
∫
a
0
f
(
a
−
x
)
d
x
, if
f
(
2
a
−
x
)
=
f
(
x
)
- note that here
log
sin
t
=
log
sin
(
π
−
t
)
and we get
I
1
=
1
2
∫
π
0
log
sin
t
d
t
=
∫
π
2
0
log
sin
t
d
t
=
I
Hence (A) becomes
2
I
=
I
−
(
π
2
)
log
2
or
I
=
−
(
π
2
)
log
2
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