integral_01

Calculus Level 3

∫ 0 π 2 ln ⁡ ( sin ⁡ x ) d x = ? \large \int _0^\frac \pi 2 \ln (\sin x) \ dx = \ ?

Round your answer to three decimal places.


The answer is -1.089.

Hassan Abdulla by Hassan Abdulla , 35, Bahrain

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2 solutions

Chew-Seong Cheong
Aug 18, 2017

Relevant wiki: Catalan's constant

I = ∫ 0 π 2 ln ⁡ ( sin ⁡ x ) d x Let 2 u = x ⟹ 2 d u = d x = 2 ∫ 0 π 4 ln ⁡ ( sin ⁡ 2 u ) d u = 2 ∫ 0 π 4 ln ⁡ ( 2 sin ⁡ u cos ⁡ u ) d u = 2 ∫ 0 π 4 ln ⁡ ( 2 sin ⁡ u ) d u + 2 ∫ 0 π 4 ln ⁡ ( cos ⁡ u ) d u = − G + 2 ∫ 0 π 4 ln ⁡ ( cos ⁡ u ) d u where G is the Catalan’s constant. = − G + 2 ∫ 0 π 4 ln ⁡ ( 2 cos ⁡ u ) d u − 2 ∫ 0 π 4 ln ⁡ 2 d u = − G + G − 2 u ln ⁡ 2 ∣ 0 π 4 = − π 2 ln ⁡ 2 ≈ − 1.089 \begin{aligned} I & = \int_0^\frac \pi 2 \ln (\sin x) \ dx & \small \color{#3D99F6} \text{Let }2u = x \implies 2du = dx \\ & = 2 \int_0^\frac \pi 4 \ln (\sin 2u) \ du \\ & = 2 \int_0^\frac \pi 4 \ln (2\sin u\cos u) \ du \\ & = {\color{#3D99F6} 2 \int_0^\frac \pi 4 \ln (2\sin u) \ du} + 2 \int_0^\frac \pi 4 \ln (\cos u) \ du \\ & = {\color{#3D99F6} -G} + 2 \int_0^\frac \pi 4 \ln (\cos u) \ du & \small \color{#3D99F6} \text{where }G \text{ is the Catalan's constant.} \\ & = -G + {\color{#3D99F6}2 \int_0^\frac \pi 4 \ln (2\cos u) \ du} - 2 \int_0^\frac \pi 4 \ln 2 \ du \\ & = -G + {\color{#3D99F6}G} - 2 u \ln 2 \ \bigg|_0^\frac \pi 4 \\ & = - \frac \pi 2 \ln 2 \approx \boxed{-1.089} \end{aligned}

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Hassan Abdulla
Aug 18, 2017

I = ∫ 0 π 2 ln ⁡ ( sin ⁡ x ) d x ⟶ 1 I\quad =\quad \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \sin { x } \right) } } dx\longrightarrow 1

I = ∫ 0 π 2 ln ⁡ ( sin ⁡ ( π 2 − x ) ) d x I\quad =\quad \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \sin { \left( \frac { \pi }{ 2 } -x \right) } \right) } } dx u = π 2 − x \textcolor{#D61F06}{\qquad u=\frac { \pi }{ 2 } -x}

I = ∫ 0 π 2 ln ⁡ ( cos ⁡ x ) d x ⟶ 2 I\quad =\quad \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \cos { x } \right) } } dx\ \longrightarrow 2 sin ⁡ ( π 2 − x ) = cos ⁡ x \textcolor{#D61F06}{\quad \sin { \left( \frac { \pi }{ 2 } -x \right) } =\cos { x } }

2 I = ∫ 0 π 2 ( ln ⁡ ( sin ⁡ x ) + ln ⁡ ( cos ⁡ x ) ) d x 2I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( \ln { \left( \sin { x } \right) } +\ln { \left( \cos { x } \right) } \right) } dx 1 + 2 \textcolor{#D61F06}{\quad 1+2 }

2 I = ∫ 0 π 2 ln ⁡ ( sin ⁡ x cos ⁡ x ) d x 2I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \sin { x } \cos { x } \right) } } dx

2 I = ∫ 0 π 2 ln ⁡ ( sin ⁡ 2 x 2 ) d x 2I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \frac { \sin { 2x } }{ 2 } \right) } } dx

2 I = ∫ 0 π 2 ( ln ⁡ ( sin ⁡ 2 x ) − ln ⁡ 2 ) d x 2I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( \ln { \left( \sin { 2x } \right) - \ln { 2 } } \right) dx }

2 I = ∫ 0 π 2 ln ⁡ ( sin ⁡ 2 x ) d x − π 2 ln ⁡ 2 2I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \sin { 2x } \right) dx-\frac { \pi }{ 2 } \ln { 2 } } }

2 I = 1 2 ∫ 0 π ln ⁡ ( sin ⁡ x ) d x − π 2 ln ⁡ 2 2I=\frac { 1 }{ 2 } \int _{ 0 }^{ \pi }{ \ln { \left( \sin { x } \right) } } dx-\frac { \pi }{ 2 } \ln { 2 } u = 2 x \textcolor{#D61F06}{\quad u=2x }

Now

∫ π 2 π ln ⁡ ( sin ⁡ x ) d x = ∫ 0 π 2 ln ⁡ ( cos ⁡ x ) d x \int _{ \frac { \pi }{ 2 } }^{ \pi }{ \ln { \left( \sin { x } \right) dx } } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \cos { x } \right) dx } } u = x − π 2 \textcolor{#D61F06}{\quad u=x-\frac { \pi }{ 2 } }

∫ 0 π 2 ln ⁡ ( cos ⁡ x ) d x = ∫ 0 π 2 ln ⁡ ( sin ⁡ x ) d x \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \cos { x } \right) } } dx=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \sin { x } \right) } } dx s e e 2 \textcolor{#D61F06}{\quad see\quad 2 }

So

2 I = 1 2 ( 2 ∫ 0 π 2 ln ⁡ ( sin ⁡ x ) d x ) − π 2 ln ⁡ 2 2I\quad =\frac { 1 }{ 2 } \left( 2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \sin { x } \right) } } dx \right) -\frac { \pi }{ 2 } \ln { 2 }

2 I = I − π 2 ln ⁡ 2 2I=I-\frac { \pi }{ 2 } \ln { 2 }

I = − π 2 ln ⁡ 2 ≈ − 1.088793 I=-\frac { \pi }{ 2 } \ln { 2 } \approx -1.088793

1 Helpful 0 Interesting 0 Brilliant 0 Confused

You did your solution very nicely, with a very nice format I might add. The only reason I was able to get this answer is because I had seen this integral before. I was watching a video entitled something like "MIT integration bee 2015" on youtube and I decided I wanted to try to solve the integral arctan(x)/x from 0 to infinity, because the integral they were given was (arctan(2x)-arctan(x))/x. The integral of ln(sin(x)) from 0 to 2pi emerged from it. At that point I just looked up the solution on the internet because I wasn't sure if the ln(sin(x))'s antiderivative was elementary or not. I was also helping someone run errands, so I wasn't able to focus on the math. But to my surprise, it was easier to figure out than I thought. I already thought of trying the double angle identity, so I know I would've figured it out. But something that I found strange was that the answer that was given during the integration bee was the same as that of the integral of arctan(x)/x. How weird is that?! I also had something happen with that integral that I hadn't seen happen before with any other integral. It totally cancelled with itself on the left and right hand sides of the equation (when integrating by parts), and gave the answer to two or three (I can't remember which) other integrals!

James Wilson - 3 years, 9 months ago

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