question_answer

Evaluate \[\int{\left[ \log (\log \,x)+\frac{1}{{{(\log \,x)}^{2}}} \right]}\,dx.\]

OR

Evaluate \[\int{\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }}\,dx.\]

Answer:

We have, \[l=\int{\left[ \log \,(\log x)+\frac{1}{{{(\log x)}^{2}}} \right]dx}\]

Put       \[{{\log }_{e}}x=z\Rightarrow x={{e}^{z}}\]

\[\therefore \]      \[dx={{e}^{z}}dz\]

Now, \[l=\int{{{e}^{z}}\left[ \log z+\frac{1}{{{z}^{2}}} \right]}\,\,dz\]

\[=\int{{{e}^{z}}\left[ \log z+\frac{1}{z}-\frac{1}{z}+\frac{1}{{{z}^{2}}} \right]}\,\,dz\]

\[\left[ \text{adding}\,\,\text{and}\,\,\text{subtracting}\frac{1}{z} \right]\]

\[=\int{{{e}^{z}}\left[ \log z+\frac{1}{z} \right]}\,\,dz-=\int{{{e}^{z}}\left[ \frac{1}{z}+\left( \frac{-\,1}{{{z}^{2}}} \right) \right]\,\,dz}\]

\[={{e}^{z}}\log z-{{e}^{z}}\cdot \frac{1}{z}+C\]

\[[\because \int{{{e}^{x}}[f\,(x)+f'(x)]\,dx={{e}^{x}}f\,(x)+C}]\]

\[={{e}^{z}}\left( \log z-\frac{1}{z} \right)+C\]

\[=x\,\left[ \log \,(\log x)-\frac{1}{\log x} \right]+C\]           \[[put\,\,z=\log x]\]

OR

We have, \[\int{\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }}\,dx\]

\[=\int{\frac{(2{{\cos }^{2}}x-1)-(2{{\cos }^{2}}\alpha -1)}{\cos x-\cos \alpha }}\,dx\]

\[=\int{\frac{2{{\cos }^{2}}x-2{{\cos }^{2}}\alpha }{\cos x-\cos \alpha }}\,dx\]

\[=\int{\frac{2\,(\cos x-\cos \alpha )\,\,(\cos x+\cos \alpha )}{(\cos x-\cos \alpha )}}\,dx\]

\[=2\int{(\cos x+\cos \alpha )}\,dx\]

\[=2\,(\sin x+x\cos \alpha )+C\]