Question

The value of $\int\limits_{\sqrt {\ln 2} }^{\sqrt {\ln 3} } {\dfrac{{x\sin {x^2}}}{{\sin {x^2} + \sin (\ln 6 - {x^2})}}dx} $ 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: Since it is a question of integration, so we will solve this question by looking at the factor with which integration part will be easy to solve for such type of integration we need to determine the result using the integration represented below, In simple words we will have to look in the integration part
\[\int\limits_a^b {f(x)dx = \int\limits_a^b {f(a + b - x)dx} } \]

Complete step by step answer:
We have to find the value of $\int\limits_{\sqrt {\ln 2} }^{\sqrt {\ln 3} } {\dfrac{{x\sin {x^2}}}{{\sin {x^2} + \sin (\ln 6 - {x^2})}}dx} $
Let \[{x^2} = t\] in the given equation
\[\Rightarrow 2xdx = dt\]
Determining the value of $xdx$, we get
\[\Rightarrow xdx = \dfrac{1}{2}dt\]
Now, After putting the required value, we get
Let $I = \dfrac{1}{2}\int\limits_{\ln 2}^{\ln 3} {\dfrac{{\sin t}}{{\sin t + \sin (\ln 6 - t)}}dt} ….....(1)$
Now using the formula,
\[\int\limits_a^b {f(x)dx = \int\limits_a^b {f(a + b - x)dx} } \]
On simplifying, we get
$\Rightarrow I = \dfrac{1}{2}\int\limits_{\ln 2}^{\ln 3} {\dfrac{{\sin [\ln 3 + \ln 2 - t]}}{{\sin (\ln 3 + \ln 2 - t) + \sin [\ln 6 - (\ln 3 + \ln 2 - t)]}}dt} $
Solving further
\[\Rightarrow I = \dfrac{1}{2}\int\limits_{\ln 2}^{\ln 3} {\dfrac{{\sin [\ln 6 - t]}}{{\sin (\ln 6 - t) + \sin [\ln 6 - \ln 6 + t]}}dt} \]
Hence, we get
\[\Rightarrow I = \dfrac{1}{2}\int\limits_{\ln 2}^{\ln 3} {\dfrac{{\sin [\ln 6 - t]}}{{\sin (\ln 6 - t) + \sin (t)}}dt} ...(2)\]
Now, adding (1) and (2), we get
$\Rightarrow 2I = \dfrac{1}{2}\int\limits_{\ln 2}^{\ln 3} {\dfrac{{\sin t}}{{\sin t + \sin (\ln 6 - t)}}dt} + \dfrac{1}{2}\int\limits_{\ln 2}^{\ln 3} {\dfrac{{\sin [\ln 6 - t]}}{{\sin (\ln 6 - t) + \sin (t)}}dt} $
So, on simplifying the above equation, we get
\[ = \dfrac{1}{2}\int\limits_{\ln 2}^{\ln 3} {\dfrac{{\sin [\ln 6 - t] + \sin t}}{{\sin (\ln 6 - t) + \sin (t)}}dt} \]
Numerator and denominator will get cancel and we get
\[\Rightarrow I = \dfrac{1}{4}\int\limits_{\ln 2}^{\ln 3} {1.dt} \]
Integrating the above equation, we get
$ = \dfrac{1}{4}[t]_{\ln 2}^{\ln 3}$
Substituting the limit, we get
$ = \dfrac{1}{4}[\ln 3 - \ln 2]$
Now, we can simplify the above equation, by using the property of logarithm
$\Rightarrow \log a - \log b = \log \dfrac{a}{b}$
Hence, by using the above equation, we get
$\Rightarrow I = \dfrac{1}{4}\ln (\dfrac{3}{2})$

$\therefore$ The value of $\int\limits_{\sqrt {\ln 2} }^{\sqrt {\ln 3} } {\dfrac{{x\sin {x^2}}}{{\sin {x^2} + \sin (\ln 6 - {x^2})}}dx} $ is $ I = \dfrac{1}{4}\ln (\dfrac{3}{2})$. Hence option (A) is correct.

Note:
In this question, carefully solve the equation after using the formula \[\int\limits_a^b {f(x)dx = \int\limits_a^b {f(a + b + x)dx} } \] and when adding both equations, don’t forget to add the left side also which will be 2I. Solve further to get the desired result.