Prove that |
\[\int_{0}^{1}{{{\sin }^{-\,1}}\left( \frac{2x}{1+{{x}^{2}}} \right)}\,dx=\frac{\pi }{2}-\log 2.\] |
OR |
Evaluate \[\int_{0}^{2}{[{{x}^{2}}]}\,dx,\] where \[[.]\]the greatest integer function is. |
Answer:
Let \[l=\int_{0}^{1}{{{\sin }^{-\,1}}\left( \frac{2x}{1+{{x}^{2}}} \right)}\,dx\] Put \[x=\tan \theta \] \[\Rightarrow \] \[dx={{\sec }^{2}}\theta d\theta \] When \[x=0,\] then \[0=ten\theta \] \[\Rightarrow \] \[\theta =0\] When \[x=1,\] then \[1=ten\,\theta \] \[\Rightarrow \] \[\theta =\pi /4\] \[\therefore \] \[l=\int_{0}^{\pi /4}{{{\sin }^{-1}}\left( \frac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)}\cdot {{\sec }^{2}}\theta \,d\theta \] \[\Rightarrow \] \[l=\int_{0}^{\pi /4}{{{\sin }^{-1}}(sin2\theta ).\,se{{c}^{2}}}\theta \,d\theta \] \[\left[ \because \,\,\,sin2\theta =\frac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right]\] \[=\int_{0}^{\pi /4}{\underset{I}{\mathop{2\theta }}\,.\underset{II}{\mathop{se{{c}^{2}}}}\,}\theta \,d\theta \] Applying the rule of integration by parts, we get \[l=\left[ 2\,\theta \int{{{\sec }^{2}}\theta \,d\theta -\int{\left\{ \frac{d}{d\theta }(2\theta )\int{{{\sec }^{2}}\theta \,d\theta } \right\}d\theta }} \right]_{0}^{\pi /4}\] \[=[2\,\theta \,\tan \theta ]_{0}^{\pi /4}-\int_{0}^{\pi /4}{2\,\tan \theta }\,d\theta \] \[=\left[ 2\times \frac{\pi }{4}\tan \frac{\pi }{4}-0 \right]-2\,[log\,sec\,\theta ]_{0}^{\pi /4}\] \[=\frac{\pi }{2}-2\left[ \log \,\sec \frac{\pi }{4}-\log \,\sec \,0 \right]\] \[=\frac{\pi }{2}-2[log\sqrt{2}-log\,1]\] \[=\frac{\pi }{2}-2.\frac{1}{2}log2=\frac{\pi }{2}-log\,2\] \[[\because \,\,\,log\,1=0]\] Hence proved. OR Here, \[[{{x}^{2}}]=\left\{ \begin{matrix} 0,\,\text{when}\,\,0\,<x<1 \\ 1,\,\text{when}\,\,1<x<\sqrt{2} \\ 2,\,\text{when}\,\,\sqrt{2}<x<\sqrt{3} \\ 3,\,\text{when}\,\,\sqrt{3}<x<2 \\ \end{matrix} \right.\] \[\therefore \,\,\int_{0}^{2}{[{{x}^{2}}]}\,dx=\int_{0}^{1}{0\,dx+\int_{1}^{\sqrt{2}}{1\,dx}}+\int_{\sqrt{2}}^{\sqrt{3}}{2\,dx}+\int_{\sqrt{3}}^{2}{3\,dx}\] \[=0+[x]_{1}^{\sqrt{2}}+[2x]_{\sqrt{2}}^{\sqrt{3}}+[3x]_{\sqrt{3}}^{2}\] \[=(\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})+3(2\,-\sqrt{3})\] \[=\sqrt{2}-1+2\sqrt{3}-2\sqrt{2}+6-3\sqrt{3}\] \[=5-\sqrt{2}-\sqrt{3}\]
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