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EXPLANATION.

[tex]\sf \displaystyle \frac{1}{3} \int_{log 3}^{log4} \frac{sin(u)}{sin(u) + sin(log 12 - u)} du[/tex]

As we know that,

King property.

[tex]\sf \displaystyle \int\limits^b_a f(x)dx = \int\limits^b_a f(a + b - x)dx[/tex]

Using this property in this question, we get.

[tex]\sf \displaystyle I = \frac{1}{3} \int_{log 3}^{log4} \frac{sin(u)}{sin(u) + sin(log 12 - u)} du. - - - - - (1)[/tex]

[tex]\sf \displaystyle I = \frac{1}{3} \int_{log 3}^{log4} \frac{sin(log 3 + log 4 - u)}{sin(log 3 + log 4 - u) + sin[log 12 - (log 3 + log 4 - u)]} du[/tex]

[tex]\sf \displaystyle I = \frac{1}{3} \int_{log 3}^{log4} \frac{sin(log 12 - u)}{sin(log 12 - u) + sin[log 12 - (log 12 - u)]} du[/tex]

[tex]\sf \displaystyle I = \frac{1}{3} \int_{log 3}^{log4} \frac{sin(log 12 - u)}{sin(log 12 - u) + sin(log 12 - log 12 + u)} du[/tex]

[tex]\sf \displaystyle I = \frac{1}{3} \int_{log 3}^{log4} \frac{sin(log 12 - u)}{sin(log 12 - u) + sin(u)} du. - - - - - (2)[/tex]

Adding expression (1) and (2), we get.

[tex]\sf \displaystyle 2I = \frac{1}{3} \int_{log 3}^{log4} \frac{sin(u) + sin(log 12 - u)}{sin (u) + sin (log 12 - u)} du[/tex]

[tex]\sf \displaystyle 2I = \frac{1}{3} \int_{log 3}^{log4} 1.du[/tex]

[tex]\sf \displaystyle 2I = \frac{1}{3} \bigg[u \bigg]_{log 3}^{log 4}[/tex]

As we know that,

In definite integration first we put upper limits then we put lower limits in the expression, we get.

[tex]\sf \displaystyle 2I = \frac{1}{3} \bigg[ log 4 - log 3 \bigg][/tex]

[tex]\sf \displaystyle I = \frac{1}{6} log \bigg(\frac{4}{3} \bigg)[/tex]

[tex]\sf \displaystyle \boxed{\frac{1}{3} \int_{log 3}^{log4} \frac{sin(u)}{sin(u) + sin(log 12 - u)} du = \frac{1}{6} log \bigg(\frac{4}{3} \bigg)}[/tex]

Answer:

[tex]\boxed{ \sf\displaystyle \frac{1}{3} \int^{ \log 4}_{ \log3} \frac{ \sin u}{ \sin u + \sin (\log12 - u)} du = \displaystyle \frac{1}{6} \log\bigg(\dfrac{4}{3} \bigg)} \\ \\ [/tex]

Step-by-step explanation:

Given integral is

[tex]\sf \: \displaystyle \frac{1}{3} \int^{ \log \: 4}_{ \log \: 3} \frac{ \sin \: u}{ \sin \: u + \sin (\log12 - u)} du \\ \\ [/tex]

Let assume that

[tex]\sf \: I = \displaystyle \frac{1}{3} \int^{ \log \: 4}_{ \log \: 3} \frac{ \sin \: u}{ \sin \: u + \sin (\log12 - u)} du - - - (1) \\ \\ [/tex]

We know,

[tex]\boxed{ \rm{ \:\displaystyle \int_{a}^{b}\sf \: f(x) \: dx \: = \: \: \displaystyle \int_{a}^{b}\sf \: f(a + b - x) \: dx}} \\\\ [/tex]

So, using this property of definite integrals, we get

[tex]\sf \: I = \displaystyle \frac{1}{3} \int^{ \log \: 4}_{ \log \: 3} \frac{ \sin( \log4 +\log3 - u)}{ \sin(\log4 + \log3 - u) + \sin (\log12 - \log4 - \log3 + u)} du \\ \\ [/tex]

[tex]\sf \: I = \displaystyle \frac{1}{3} \int^{ \log \: 4}_{ \log \: 3} \frac{ \sin( \log12 - u)}{ \sin(\log12 - u) + \sin (\log12 - \log12 + u)} du \\ \\ [/tex]

[tex]\sf \: I = \displaystyle \frac{1}{3} \int^{ \log \: 4}_{ \log \: 3} \frac{ \sin( \log12 - u)}{ \sin(\log12 - u) + \sin ( u)} du - - - (2)\\ \\ [/tex]

On adding equation (1) and (2), we get

[tex]\sf \:2I = \displaystyle \frac{1}{3} \int^{ \log \: 4}_{ \log \: 3} \frac{ \sin(u) + \sin( \log12 - u)}{ \sin(\log12 - u) + \sin ( u)} du \\ \\ [/tex]

[tex]\sf \:2I = \displaystyle \frac{1}{3} \int^{ \log \: 4}_{ \log \: 3} du \\ \\ [/tex]

[tex]\sf \:2I = \displaystyle \frac{1}{3} \bigg(u\bigg) ^{ \log \: 4}_{ \log \: 3} \\ \\ [/tex]

[tex]\sf \:2I = \displaystyle \frac{1}{3} \bigg(\log4 - \log3\bigg) \\ \\ [/tex]

[tex]\sf \:2I = \displaystyle \frac{1}{3} \log\bigg(\dfrac{4}{3} \bigg) \\ \\ [/tex]

[tex]\sf \:I = \displaystyle \frac{1}{6} \log\bigg(\dfrac{4}{3} \bigg) \\ \\ [/tex]

Hence,

[tex]\sf \:\displaystyle \frac{1}{3} \int^{ \log \: 4}_{ \log \: 3} \frac{ \sin \: u}{ \sin \: u + \sin (\log12 - u)} du = \displaystyle \frac{1}{6} \log\bigg(\dfrac{4}{3} \bigg) \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

[tex] {{\bf{Additional\:Information}}}[/tex]

[tex]\boxed{ \rm{ \:\displaystyle \int_{a}^{b}\sf \: f(x) \: dx \: = - \: \: \displaystyle \int_{b}^{a}\sf \: f(x) \: dx}} \\ [/tex]

[tex]\boxed{ \rm{ \:\displaystyle \int_{a}^{b}\sf \: f(x) \: dx \: = \: \: \displaystyle \int_{a}^{b}\sf \: f(y) \: dy}} \\ [/tex]

[tex]\boxed{ \rm{ \:\displaystyle \int_{0}^{a}\sf \: f(x) \: dx \: = \: \: \displaystyle \int_{0}^{a}\sf \: f(a - x) \: dx}} \\ [/tex]

[tex]\boxed{ \rm{ \:\displaystyle \int_{a}^{b}\sf \: f(x) \: dx \: = \: \: \displaystyle \int_{a}^{b}\sf \: f(a + b - x) \: dx}} \\ [/tex]

[tex]\boxed{ \rm{ \:\displaystyle \int_{0}^{2a}\sf \: f(x) \: dx \: = \:2 \displaystyle \int_{0}^{2}\sf f(x)dx \: \: if \: f(2a - x) = f(x)}} \\ [/tex]

[tex]\boxed{ \rm{ \:\displaystyle \int_{0}^{2a}\sf \: f(x) \: dx \: = \:0 \: \: if \: f(2a - x) = - \: f(x)}} \\ [/tex]