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