To 3 decimal places, evaluate
∫ 0 π / 2 ln ( sin ( x ) ) d x .
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how is ln sinx and ln cos x equal that u'hv added these two?
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the area under the curves from 0 to pi/2 is symmetric
got it from suddep's explanation :)
I also used the exact approach but got it wrong at the end
Using properties of definite integrals it can easily be evaluated. I will use the following properties:
0 ∫ a f ( x ) d x = 0 ∫ a f ( a − x ) d x
0 ∫ 2 a f ( x ) d x = 2 × 0 ∫ a f ( x ) d x , if f ( x ) = f ( 2 a − x )
a ∫ b f ( x ) d x = a ∫ b f ( t ) d t
Let I = 0 ∫ 2 π l n ( sin x ) d x ⇒ I = 0 ∫ 2 π l n ( cos x ) d x
⇒ 2 I = 0 ∫ 2 π ( l n ( sin x ) + l n ( cos x ) ) d x ⇒ 2 I = 0 ∫ 2 π l n ( sin x cos x ) d x
⇒ 2 I = 0 ∫ 2 π l n ( sin 2 x ) d x − 0 ∫ 2 π l n ( 2 ) d x
Put
2
x
=
t
to get,
2
I
=
2
1
0
∫
π
l
n
(
sin
t
)
d
t
−
2
π
l
n
2
⇒ 2 I = 2 2 0 ∫ 2 π l n ( sin t ) d t − 2 π l n 2
⇒ 2 I = I − 2 π l n 2
Therefore,
0
∫
2
π
l
n
(
sin
x
)
d
x
=
−
2
π
l
n
2
=
−
1
.
0
8
8
7
9
You wouldn't believe but we wrote the solutions at nearly identical times!
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Not only that but our solutions are same word to word too. :D
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I = ∫ 0 π / 2 l n ( s i n x ) d x
I = ∫ 0 π / 2 l n ( c o s x ) d x
Adding,
2 I = ∫ 0 π / 2 l n ( 2 s i n 2 x ) d x
2 I = ∫ 0 π / 2 l n ( s i n 2 x ) − l n 2 d x
2 I = ∫ 0 π / 2 l n ( s i n 2 x ) d x − 2 π l n 2
Let 2 x = t . Then, 2 d x = d t
2 I = 2 1 ∫ 0 π l n ( s i n t ) d t − 2 π l n 2
2 I = ∫ 0 π / 2 l n ( s i n t ) d t − 2 π l n 2
2 I = I − 2 π l n 2
I = − 2 π l n 2
I = − 1 . 0 8 9