Question

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

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

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

Let assume that

$\rm :\longmapsto\:I = \displaystyle \int^{ \sqrt[3]{ \log4} }_{ \sqrt[3]{ \log3} } \frac{ {x}^{2} \sin {x}^{3} }{ \sin {x}^{3} + \sin( \log12 - {x}^{3} )} \: dx$

To, solve this integral, we use Method of Substitution.

So, Substitute

$\red{\rm :\longmapsto\: {x}^{3} = y}$

$\red{\rm :\longmapsto\: 3{x}^{2}dx = dy}$

$\red{\rm :\longmapsto\: {x}^{2}dx = \dfrac{dy}{3} }$

Now, we know, when we substitute, we have to change the limits.

So,

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

So, above integral can be rewritten as

$\rm :\longmapsto\:I =\dfrac{1}{3} \displaystyle \int^{log4 }_{ log3 } \frac{\sin y }{ \sin y + \sin( \log12 - y )} \: dy - - (1)$

We know

$\boxed{\tt{ \displaystyle\int_a^b \: f(x)dx = \displaystyle\int_a^b \: f(a + b - x) \: dx \: }}$

So, using this identity, Change

$\red{\rm :\longmapsto\:x \to \: log4 + log3 - x = log12 - x}$

So, we get

$\rm :\longmapsto\:I =\dfrac{1}{3} \displaystyle \int^{log4 }_{ log3 } \frac{\sin (log12 - y) }{ \sin (log12 - y) + \sin(y)} \: dy - - (2)$

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

$\rm :\longmapsto\:2I =\dfrac{1}{3} \displaystyle \int^{log4 }_{ log3 } \frac{sin(y) + \sin (log12 - y) }{ \sin (log12 - y) + \sin(y)} \: dy$

$\rm :\longmapsto\:2I =\dfrac{1}{3} \displaystyle \int^{log4 }_{ log3 } 1 \: dy$

$\rm :\longmapsto\:2I =\dfrac{1}{3} \bigg[y\bigg]^{log4 }_{ log3 }$

$\rm :\longmapsto\:I =\dfrac{1}{6} \bigg[{log4 } - { log3 }\bigg]$

$\rm :\longmapsto\:I =\dfrac{1}{6} \bigg[{log \dfrac{4}{3} }\bigg]$

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More to know :-

$\boxed{\tt{ \displaystyle\int _a^b \: f(x) \: dx \: = \: \displaystyle\int _a^bf(y) \: dy \: }}$

$\boxed{\tt{ \displaystyle\int _a^b \: f(x) \: dx \: = \: - \: \displaystyle\int _b^a \: f(x) \: dx \: }}$

$\boxed{\tt{ \displaystyle\int _0^a \: f(x) \: dx \: = \: \displaystyle\int _0^a \: f(a - x) \: dx \: }}$

$\boxed{\tt{ \displaystyle\int _a^b \: f(x) \: dx \: = \: \displaystyle\int _a^b \: f(a + b - x) \: dx \: }}$

$\boxed{\tt{ \displaystyle\int _{ - a}^a\: f(x) \: dx \: = 0 \: \: if \: f( - x) = - f(x)}}$

$\boxed{\tt{ \displaystyle\int _{ - a}^a\: f(x) \: dx \: = 2\displaystyle\int _0^af(x)dx \: \: if \: f( - x) = f(x)}}$