question_answer

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,     \[[{{x}^{2}}]=\left\{ \begin{matrix}    0,\,\text{when}\,\,0\,<x<1  \\    1,\,\text{when}\,\,1<x<\sqrt{2}  \\    2,\,\text{when}\,\,\sqrt{2}<x<\sqrt{3}  \\    3,\,\text{when}\,\,\sqrt{3}<x<2  \\ \end{matrix} \right.\]

\[\therefore \,\,\int_{0}^{2}{[{{x}^{2}}]}\,dx=\int_{0}^{1}{0\,dx+\int_{1}^{\sqrt{2}}{1\,dx}}+\int_{\sqrt{2}}^{\sqrt{3}}{2\,dx}+\int_{\sqrt{3}}^{2}{3\,dx}\]            \[=0+[x]_{1}^{\sqrt{2}}+[2x]_{\sqrt{2}}^{\sqrt{3}}+[3x]_{\sqrt{3}}^{2}\]

\[=(\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})+3(2\,-\sqrt{3})\]

\[=\sqrt{2}-1+2\sqrt{3}-2\sqrt{2}+6-3\sqrt{3}\]

\[=5-\sqrt{2}-\sqrt{3}\]