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[tex]\sf \: \displaystyle \:\int^{ \sqrt[3]{ \log4} }_ {\sqrt[3]{ \log3}} \frac{ {x}^{2} \sin \: {x}^{3} }{ \sin \: {x}^{3} + \sin (\log12 - {x}^{3}) } dx \\ \\ [/tex]

Answer:

[tex]\boxed{ \sf{ \: \displaystyle\int^{ \sqrt[3]{ \log4} }_ {\sqrt[3]{ \log3}} \sf \frac{ {x}^{2} \sin {x}^{3} }{ \sin {x}^{3} + \sin (\log12 - {x}^{3}) } dx = \frac{1}{6} \: log\bigg(\dfrac{4}{3} \bigg) \: }} \\ \\ [/tex]

Step-by-step explanation:

Given integral is

[tex]\sf \: \displaystyle \:\int^{ \sqrt[3]{ \log 4} }_ {\sqrt[3]{ \log3}} \frac{ {x}^{2} \sin \: {x}^{3} }{ \sin \: {x}^{3} + \sin (\log12 - {x}^{3}) } dx \\ \\ [/tex]

can be rewritten as

[tex]\sf \:I = \frac{1}{3} \displaystyle \:\int^{ \sqrt[3]{ \log4} }_ {\sqrt[3]{ \log 3}} \frac{3 {x}^{2} \sin \: {x}^{3} }{ \sin \: {x}^{3} + \sin (\log12 - {x}^{3}) } dx \\ \\ [/tex]

To solve this integral further, we use method of substitution.

So, Substitute

[tex]\sf \: {x}^{3} = y \\ \\ [/tex]

[tex]\sf \: 3{x}^{2} \: dx \: = \: dy \\ \\ [/tex]

In definite integral, when we substitute, we have to change limits too.

So,

[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf \sqrt[3]{log3} & \sf log3 \\ \\ \sf \sqrt[3]{log4} & \sf log4 \end{array}} \\ \end{gathered} \\ \\ [/tex]

So, using these, the above integral can be rewritten as

[tex]\sf \:I = \frac{1}{3} \displaystyle \:\int^{log4 }_ {log3} \frac{ \sin \: y }{ \sin \: y + \sin (\log12 - y) } dy - - - (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 have

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

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

[tex]\qquad \boxed{ \sf{ \: \because \: logx + logy = log(xy) \: }} \\ \\ [/tex]

[tex]\sf \:I = \frac{1}{3} \displaystyle \:\int^{log4 }_ {log3} \frac{ \sin \: (log12 - y) }{ \sin (log12 - y) + \sin y } dy - - - (2) \\ \\ [/tex]

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

[tex]\sf \:2I = \frac{1}{3} \displaystyle \:\int^{log4 }_ {log3} \frac{ \sin \: (log12 - y) + \sin y}{ \sin (log12 - y) + \sin y } dy \\ \\ [/tex]

[tex]\sf \:2I = \frac{1}{3} \displaystyle \:\int^{log4 }_ {log3} dy \\ \\ [/tex]

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

[tex]\sf \:I = \frac{1}{6} \bigg(log4 - log3\bigg) \\ \\ [/tex]

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

Hence,

[tex]\sf\sf\implies \displaystyle\int^{ \sqrt[3]{ \log4} }_ {\sqrt[3]{ \log3}} \sf \frac{ {x}^{2} \sin {x}^{3} }{ \sin {x}^{3} + \sin (\log12 - {x}^{3}) } dx = \frac{1}{6} log\bigg(\dfrac{4}{3} \bigg) \\ \\ [/tex]

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

Additional Information

[tex] { \red{ \mathfrak{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]