question_answer

Evaluate \[\int_{0}^{2\pi }{\frac{x{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}}\,dx\]

OR

For x > 0, let \[f(x)=\int_{1}^{x}{\frac{{{\log }_{e}}t}{1+t}}\,dt.\] Find function \[f(x)+f\left( \frac{1}{x} \right)\] and show that \[f(e)+f\left( \frac{1}{e} \right)=\frac{1}{2}.\]

Answer:

Let \[I=\int_{0}^{2\pi }{\frac{x\,{{\sin }^{2n}}}{{{\sin }^{2n}}+{{\cos }^{2n}}x}}\,dx\]                  ?(i)

Then, \[I=\int_{0}^{2\pi }{\frac{(2\pi -x)\,{{\sin }^{2n}}(2\pi -x)}{{{\sin }^{2n}}(2\pi -x)+{{\cos }^{2n}}(2\pi -x)}}\,dx\]

\[\Rightarrow \]   \[I=\int_{0}^{2\pi }{\frac{(2\pi -x)\,{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}}\,dx\]

On adding Eqs. (i) and (ii), we get

\[2I=\int_{0}^{2\pi }{\frac{2\pi \,{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}}\,dx\]

\[\Rightarrow \]   \[I=\pi \int_{0}^{2\pi }{\frac{{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}}\,dx\]

\[\Rightarrow \]   \[I=2\pi \int_{0}^{\pi }{\frac{{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}}\,dx\]

\[\left[ \because \int_{0}^{2a}{f(x)dx}=2\int_{0}^{a}{f(x),\,\,if\,\,f(2a-x)=f(x)} \right]\]\[\Rightarrow \]            \[I=4\pi \int_{0}^{\pi /2}{\frac{{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}}\,dx\]          ?(iii)

\[\Rightarrow \] \[I=4\pi \int_{0}^{\pi /2}{\frac{{{\sin }^{2n}}(\pi /2-x)}{{{\sin }^{2n}}(\pi /2-x)+{{\cos }^{2n}}(\pi /2-x)}}\,dx\]

\[\Rightarrow \]   \[I=4\pi \int_{0}^{\pi /2}{\frac{{{\cos }^{2n}}x}{{{\cos }^{2n}}x+{{\sin }^{2n}}x}}\,dx\]            ?(iv)

On adding Eqs. (iii) and (iv), we get

\[2l=4\pi \int_{0}^{\pi /2}{\left[ \frac{{{\sin }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x}+\frac{{{\cos }^{2n}}x}{{{\sin }^{2n}}x+{{\cos }^{2n}}x} \right]}\,dx\]    \[=\int_{0}^{\pi /2}{1\,dx=4\pi \times \frac{\pi }{2}}\]

\[\therefore \,\,\,I={{\pi }^{2}}\]

OR

Given, \[f(x)=\int_{1}^{x}{\frac{{{\log }_{e}}t}{1+t}\,dt}\]                                  ?(i)

Then, \[f\left( \frac{1}{x} \right)=\int_{1}^{1/x}{\frac{{{\log }_{e}}t}{1+t}\,dt}\]

Put \[t=\frac{1}{u}\] \[\Rightarrow \] \[dt=-\frac{1}{{{u}^{2}}}du\]

Also, if t = 1, then u = 1 and if \[t=\frac{1}{x},\] then u=x

\[\therefore \] \[t\left( \frac{1}{x} \right)=\int_{1}^{x}{\frac{{{\log }_{e}}\frac{1}{u}}{1+\frac{1}{u}}\times \frac{(-\,1)}{{{u}^{2}}}}\,du\]

\[=\int_{1}^{x}{\frac{{{\log }_{e}}u}{(1+u)u}}\,du\]

\[\left[ \because \,\,\,{{\log }_{e}}\frac{1}{u}={{\log }_{e}}1-{{\log }_{e}}u=0-{{\log }_{e}}u \right]\]

\[=\int_{1}^{x}{\frac{{{\log }_{e}}t}{(1+t)t}}\,dt\]       [replace u by t]   ?(ii)

On adding Eqs. (i) and (ii), we get

\[t(x)+f\left( \frac{1}{x} \right)=\int_{1}^{x}{\left\{ \frac{{{\log }_{e}}t}{1+t}+\frac{{{\log }_{e}}t}{(1+t)t} \right\}}\,dt\]

\[=\int_{1}^{x}{\frac{{{\log }_{e}}t}{1+t}\left( \frac{1+t}{t} \right)}\,dt\]

\[=\int_{1}^{x}{\frac{{{\log }_{e}}t}{t}}\,dt\]                          ?(iii)

Again, put \[=v={{\log }_{e}}t\] \[\Rightarrow \] \[\frac{1}{t}dt=dv\]

Also, if t = x, then \[v={{\log }_{e}}x\] and if t = 1, then

v = 0 then, from Eq. (iii), we get

\[f(x)+f\left( \frac{1}{x} \right)=\int_{0}^{{{\log }_{e}}x}{v\,d}v=\left[ \frac{{{v}^{2}}}{2} \right]_{0}^{{{\log }_{e}}x}=\frac{1}{2}{{[{{\log }_{e}}x]}^{2}}-0\]  At\[x=e,f(e)+f\left( \frac{1}{e} \right)=\frac{1}{2}{{({{\log }_{e}}e)}^{2}}=\frac{1}{2}\]

\[[\because \,{{\log }_{a}}a=1]\]