Question

The value of $ \int_{\sqrt{\ln 2}}^{\sqrt{\ln 3}}{\dfrac{x\sin {{x}^{2}}}{\sin {{x}^{2}}+\sin \left( \ln 6-{{x}^{2}} \right)}} $ is \[\]
A. $ \dfrac{1}{4}\ln \dfrac{3}{2} $ \[\]
B. $ \dfrac{1}{2}\ln \dfrac{3}{2} $ \[\]
C. $ \ln \dfrac{3}{2} $ \[\]
D. $ \dfrac{1}{6}\ln \dfrac{3}{2} $ \[\]

Answer 1

Hint: We use change of variable method and take $ u={{x}^{2}} $ which we differentiate with respect to $ x $ to get $ xdx=\dfrac{du}{2} $ . We change the limits of definite integral for the new variable $ u $ accordingly and the use following property of definite integration $ \int_{a}^{b}{f\left( x \right)dx}=\int_{a}^{b}{f\left( a+b-x \right)dx} $ . \[\]

Complete step by step answer:
We know that anti-derivative, primitive function or indefinite integral of a function $ f $ is a differentiable function $ F $ whose derivative is equal to the original function $ f $ which means $ {{F}^{'}}=f $ . The process of finding integral is called integration and the original function is called integrand. We write integration in variable $ x $ as
\[\int{f\left( x \right)}dx=F\left( x \right)+c\]
 Here $ c $ is an arbitrary real constant of integration. We also integral remains same even if we change the variable. It means for any variable $ u $ we have;
\[\int{f\left( u \right)}du=\int{f\left( u \right)}du\]
If we have composite function $ f\left( g\left( x \right) \right) $ and the differential of the function inside the bracket $ {{g}^{'}}\left( x \right) $ we can substitute the $ g\left( x \right) $ as by variable $ u $ which means $ g\left( x \right)=u $ we can integrate as
\[\int{f\left( g\left( x \right) \right){{g}^{'}}\left( x \right)}dx=\int{f\left( u \right)}du\]

The above method is called integration by substitution, u-substitution or change of variable method. When we integrate under within a limit $ \left[ a,b \right] $ we get the definite integral, \[\int_{a}^{b}{f\left( x \right)dx}=\left[ F\left( x \right) \right]_{a}^{b}=F\left( b \right)-F\left( a \right)\]
 We also know the following identity of definite integration
\[\int_{a}^{b}{f\left( x \right)dx}=\int_{a}^{b}{f\left( a+b-x \right)dx}\]
We are asked in the question to evaluate following integral.
\[\begin{align}
  & \int_{\sqrt{\ln 2}}^{\sqrt{\ln 3}}{\dfrac{x\sin {{x}^{2}}}{\sin {{x}^{2}}+\sin \left( \ln 6-{{x}^{2}} \right)}}dx \\
 & \Rightarrow \int_{\sqrt{\ln 2}}^{\sqrt{\ln 3}}{\dfrac{\sin {{x}^{2}}\times xdx}{\sin {{x}^{2}}+\sin \left( \ln 6-{{x}^{2}} \right)}} \\
\end{align}\]
We see in the integrand that $ {{x}^{2}} $ is repeated in both numerator and denominator and we know that $ \dfrac{d}{dx}{{x}^{2}}=2x $ and $ x $ is also present numerator. So let use u-substitution method an take \[u={{x}^{2}}\]
We differentiate both side with respect to $ x $ and have;
\[\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( u \right)=\dfrac{d}{dx}\left( {{x}^{2}} \right) \\
 & \Rightarrow \dfrac{du}{dx}=2x \\
 & \Rightarrow du=2xdx \\
 & \Rightarrow xdx=\dfrac{du}{2} \\
\end{align}\]
We also have to change the limits for new variable. So we have
\[\begin{align}
  & x=\sqrt{\ln 2}\Rightarrow u={{x}^{2}}={{\left( \sqrt{\ln 2} \right)}^{2}}=\ln 2 \\
 & x=\sqrt{\ln 3}\Rightarrow u={{x}^{2}}={{\left( \sqrt{\ln 3} \right)}^{2}}=\ln 3 \\
\end{align}\]
We replace the limits , the variable $ {{x}^{2}}=u $ and $ xdx=\dfrac{du}{2} $ in the integrand to have;
\[\begin{align}
  & \Rightarrow \int_{\ln 2}^{\ln 3}{\dfrac{\sin u\times \dfrac{du}{2}}{\sin u+\sin \left( \ln 6-u \right)}} \\
 & \Rightarrow \dfrac{1}{2}\int_{\ln 2}^{\ln 3}{\dfrac{\sin u}{\sin u+\sin \left( \ln 6-u \right)}}du=I\left( \text{say} \right).......\left( 1 \right) \\
\end{align}\]
We use the property of definite integration \[\int_{a}^{b}{f\left( u \right)dx}=\int_{a}^{b}{f\left( a+b-u \right)du}\] for $ a=\ln 2,b=\ln 3 $ since $ a+b=\ln 2+\ln 3=\ln \left( 2\times 3 \right)=\ln 6 $ and $ f\left( u \right)=\sin u $ in the above step to have;
\[\begin{align}
  & \Rightarrow I=\dfrac{1}{2}\int_{\ln 2}^{\ln 3}{\dfrac{\sin \left( \ln 6-u \right)}{\sin \left( \ln 6-u \right)+\sin \left( \ln 6-\left( \ln 6-u \right) \right)}}du \\
 & \Rightarrow I=\dfrac{1}{2}\int_{\ln 2}^{\ln 3}{\dfrac{\sin \left( \ln 6-u \right)}{\sin \left( \ln 6-u \right)+\sin \left( u \right)}}du.......\left( 2 \right) \\
\end{align}\]
We add respective sides of equation (1) and (2) to have;
\[\begin{align}
  & 2I=\dfrac{1}{2}\int_{\ln 2}^{\ln 3}{\dfrac{\sin u}{\sin u+\sin \left( \ln 6-u \right)}}du+\dfrac{1}{2}\int_{\ln 2}^{\ln 3}{\dfrac{\sin \left( \ln 6-u \right)}{\sin \left( \ln 6-u \right)+\sin \left( u \right)}}du \\
 & \Rightarrow 2I=\dfrac{1}{2}\int_{\ln 2}^{\ln 3}{\dfrac{\sin u+\sin \left( \ln 6-u \right)}{\sin u+\sin \left( \ln 6-u \right)}du} \\
 & \Rightarrow I=\dfrac{1}{4}\int_{\ln 2}^{\ln 3}{1\cdot du} \\
 & \Rightarrow I=\dfrac{1}{4}\left[ u \right]_{\ln 2}^{\ln 3}=\dfrac{1}{4}\left( \ln 3-\ln 2 \right)=\dfrac{1}{4}\ln \left( \dfrac{3}{2} \right) \\
\end{align}\]
So the correct option is A. \[\]

Note:
We have used the logarithmic identity of product $ \ln \left( ab \right)=\ln a+\ln b $ and logarithmic identity of quotient $ \ln \left( \dfrac{a}{b} \right)=\ln a-\ln b $ for $ a=\ln 3,b=\ln 2 $ in the solution. We must not forget to change the limits after we substitute the new variable $ u $ . We note that definite integration $ \int_{a}^{b}{f\left( x \right)} $ results in always a positive number and represents the area under the curve in the interval $ \left[ a,b \right] $ .