Question

Find the value of \[\int {\dfrac{{\cos \,x + x\,\sin \,x}}{{{x^2} + x\,\cos \,x}}} dx\].
A. \[\log \left| {\dfrac{{\sin x}}{{1 + \cos x}}} \right| + c\]
B. \[\log \left| {\dfrac{{\sin x}}{{x + \cos x}}} \right| + c\]
C. \[\log \left| {\dfrac{{2\sin x}}{{x + \cos x}}} \right| + c\]
D. \[\log \left| {\dfrac{{x\sin x}}{{x + \cos x}}} \right| + c\]
E. \[\log \left| {\dfrac{x}{{x + \cos x}}} \right| + c\]

Answer 1

Hint: We will break the given integration into two parts. In the first part, the numerator is a factor of the denominator. The second part will be such that the derivative of the denominator is the numerator. Then we will integrate both integrations to get the required value.

Formula Used:
Integration formula
\[\int {\dfrac{1}{x}} dx = \log \left| x \right| + c\]
Derivative formula
Power rule \[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\]
Derivative of cosine function\[\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x\]
Quotient law for logarithm: \[\log a - \log b = \log \dfrac{a}{b}\]
Complete step by step solution:
Given integration is \[\int {\dfrac{{\cos \,x + x\,\sin \,x}}{{{x^2} + x\,\cos \,x}}} dx\]
Let \[I = \int {\dfrac{{\cos \,x + x\,\sin \,x}}{{{x^2} + x\,\cos \,x}}} dx\]
Now add and subtract \[x\] with the numerator.
\[I = \int {\dfrac{{\cos \,x + x\,\sin \,x + x - x}}{{{x^2} + x\,\cos \,x}}} dx\]
Rewrite the numerator as sum of two expressions
\[I = \int {\dfrac{{\left( {\cos \,x + x} \right) + \left( {x\,\sin \,x - x} \right)}}{{{x^2} + x\,\cos \,x}}} dx\]
Divide the integration into two integrations.
\[I = \int {\dfrac{{\cos \,x + x}}{{{x^2} + x\,\cos \,x}}} dx + \int {\dfrac{{x\,\sin \,x - x}}{{{x^2} + x\,\cos \,x}}dx} \]
Simplify the first and second integrations
\[I = \int {\dfrac{{\cos \,x + x}}{{x\left( {\cos \,x + x} \right)}}} dx + \int {\dfrac{{x\,\left( {\sin \,x - 1} \right)}}{{x\left( {x + \,\cos \,x} \right)}}dx} \]
Cancel out \[\left( {\cos \,x + x} \right)\] from the first integration and \[x\] from the second integration
\[ \Rightarrow I = \int {\dfrac{1}{x}} dx + \int {\dfrac{{\sin x - 1}}{{x + \cos \,x}}dx} \]
Now let \[{I_1} = \int {\dfrac{1}{x}} dx\] and \[{I_2} = \int {\dfrac{{\sin x - 1}}{{x + \cos \,x}}dx} \]
So, \[I = {I_1} + {I_2}\] …(i)
Apply the formula \[\int {\dfrac{1}{x}} dx = \log \left| x \right| + c\] in \[{I_1}\]
\[{I_1} = \int {\dfrac{1}{x}} dx\]
\[ \Rightarrow {I_1} = \log \left| x \right| + {c_1}\] where \[{c_1}\]is an integration constant. …….(ii)
Now we will calculate the value of \[{I_2}\].
\[{I_2} = \int {\dfrac{{\sin x - 1}}{{x + \cos \,x}}dx} \]
Assume that \[x + \cos x = z\]
Differentiate both sides of the above equation
\[\dfrac{d}{{dx}}\left( {x + \cos x} \right) = \dfrac{d}{{dx}}\left( z \right)\]
Break the left side
\[\dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( {\cos x} \right) = \dfrac{d}{{dx}}\left( z \right)\]

Apply the power rule \[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\] and derivative of cosine function\[\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x\]
\[1 - \sin x = \dfrac{{dz}}{{dx}}\]
\[ \Rightarrow \left( {1 - \sin x} \right)dx = dz\]
So, \[{I_2} = \int {\dfrac{{\sin x - 1}}{{x + \cos \,x}}dx} \] becomes after substituting \[x + \cos x = z\] and \[\left( {1 - \sin x} \right)dx = dz\]
\[{I_2} = - \int {\dfrac{1}{z}dz} \]
Applying the integration formula \[\int {\dfrac{1}{x}} dx = \log \left| x \right| + c\]
\[{I_2} = - \log \left| z \right| + {c_2}\] where \[{c_2}\]is an integration constant.
Putting the value of \[z\]
\[{I_2} = - \log \left| {x + \cos x} \right| + {c_2}\] …….(iii)
Now putting the value of \[{I_1}\] and \[{I_2}\] in equation (i)
\[I = \log \left| x \right| - \log \left| {x + \cos x} \right| + c\] where \[{c_1} + {c_2} = c\].
Apply the logarithm formula \[\log a - \log b = \log \dfrac{a}{b}\]
\[I = \log \left| {\dfrac{x}{{x + \cos x}}} \right| + c\]

Hence option E is the correct option.

Note: Students often do mistake to solve this type of integration. Instead of adding and subtracting \[x\] with numerator, they directly break it into two parts like \[\int {\dfrac{{\cos \,x + x\,\sin \,x}}{{{x^2} + x\,\cos \,x}}} dx = \int {\dfrac{{\cos \,x}}{{{x^2} + x\,\cos \,x}}} dx + \int {\dfrac{{x\,\sin \,x}}{{{x^2} + x\,\cos \,x}}} dx\] . They are stuck in this place, then unable to solve the integration. Sometimes they forgot the rule of logarithm and were unable to reach the final answer.