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Care to elaborate on your solution? Currently as it is, the solution is very incomplete and doesn't help in understanding how the integral is evaluated.
You missed mentioning many things like your use of IBP (note that ILATE is just a mnemonic that helps when you use IBP, it isn't a rule by itself). You also need to at least give a brief explanation of how the simpler integrals and expressions are evaluated to get to the final answer.
Also, it needs to be mentioned in the solution that this is an improper definite integral which would explain why we do the calculation taking limits at endpoints.
Surely at least one and possibly both terms in your solution are undefined (ln(0))?
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Take the limit x → 0 + lim x ⋅ lo g ( sin x ) = 0
I can't understand,
its "ln" not "log"
Using integration by parts with u = x and d v = cot x d x , we have: I = x ln ( sin x ) ∣ 0 π / 2 − ∫ 0 π / 2 ln ( sin x ) d x
Let I 1 = ∫ 0 π / 2 ln ( sin x ) d x , and note that:
I 1 = ∫ 0 π / 2 ln ( 2 sin ( 2 x ) cos ( 2 x ) d x = ∫ 0 π / 2 ln ( 2 ) d x + ∫ 0 π / 2 ln ( sin ( 2 x ) ) d x + ∫ 0 π / 2 ln ( cos ( 2 x ) ) d x = 2 π l n ( 2 ) + I 2 + I 3
Let x = 2 u so that I 2 + I 3 = 2 ∫ 0 π / 4 ln ( sin u ) d u + 2 ∫ 0 π / 4 ln ( cos u ) d u
Let u = 2 π − t so that 2 ∫ 0 π / 4 ln ( cos u ) d u = 2 ∫ π / 4 π / 2 ln ( sin t ) d t , and so I 1 = 2 π l n ( 2 ) + 2 ∫ 0 π / 2 ln ( sin x ) d x = 2 π l n ( 2 ) + 2 I 1 I 1 = − 2 π l n ( 2 )
Finally, I = b → 0 lim b ln ( sin b ) − ( − 2 π l n ( 2 ) ) = 0 + 2 π l n ( 2 ) ∼ 1 . 0 8 8 , where the limit is evaluated using L’Hospital’s Rule.
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ILATE RULE
x lo g ( sin x ) ∣ ∣ ∣ ∣ 0 π / 2 − 0 ∫ π / 2 lo g ( sin x ) d x = − ( − 2 π lo g 2 ) = 1 . 0 8 8 7 9