
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
477k+ views
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.
\[\int\limits_a^b {f(x)dx = \int\limits_a^b {f(a + b - x)dx} } \]
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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

What is a transformer Explain the principle construction class 12 physics CBSE

Who is Mukesh What is his dream Why does it look like class 12 english CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

What are the major means of transport Explain each class 12 social science CBSE
