question_answer

Evaluate \[\int_{0}^{\pi /2}{\log (\sin \,x)\,dx.}\]

OR

Evaluate \[\int{\frac{{{\sin }^{6}}x+{{\cos }^{6}}x}{{{\sin }^{2}}{{\cos }^{2}}x}}\,dx.\]

Answer:

Let        \[l=\int\limits_{0}^{\pi /2}{\log \,\,(\sin x)\,\,dx}\]

\[\therefore \]      \[l=\int\limits_{0}^{\pi /2}{\log \,\,\left( \sin \left( \frac{\pi }{2}-x \right) \right)\,\,dx}\]

\[=\int\limits_{0}^{\pi /2}{\log \,\,(\cos x)\,\,dx}\]              ?(ii)

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

\[2l=\int\limits_{0}^{\pi /2}{\{\log \,\,(\sin x)+\log \,\,(\cos x)\}\,\,dx}\]

\[=\int\limits_{0}^{\pi /2}{(\log \,\,(\sin x\cdot \cos x))\,\,dx}\]

\[=\int\limits_{0}^{\pi /2}{(\log \,\,\left( \frac{2\sin x\cos x}{2} \right)\,\,dx}\]

\[\Rightarrow \]\[2l=\int\limits_{0}^{\pi /2}{(\log \,\,(\sin 2x)\,\,dx-\int\limits_{0}^{\pi /2}{\log \,\,2\,\,dx}}\]  ?(iii)

Put       \[2x=t\Rightarrow 2dx=dt\Rightarrow dx=\frac{1}{2}dt\]

When    \[x=0,\]then \[t=0\]and \[x=\frac{\pi }{2},\]then \[t=\pi \]

From Eq. (iii),

Now,\[\int\limits_{0}^{\pi /2}{\log \,\,(\sin 2x)\,\,dx}=\int\limits_{0}^{\pi }{\frac{1}{2}(\log \,\,(\sin t))\,\,dt}\]

\[=\frac{2}{2}\int\limits_{0}^{\pi /2}{[\log \,\,(\sin t)]\,\,dt}=\int\limits_{0}^{\pi }{\frac{1}{2}\log \,\,(\sin t)\,\,dt=l}\]

From Eq. (iii), we get

\[2l=l-\int\limits_{0}^{\pi /2}{\log \,\,2\,\,dx}\]

\[\Rightarrow \]   \[l=[-\log \,\,(2)\,\,x]_{0}^{\pi /2}=-\frac{\pi }{2}\log 2\]

OR

Let

\[l=\int{\frac{{{\sin }^{6}}x+{{\cos }^{6}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}\,\,}dx=\int{\frac{{{({{\sin }^{2}}x)}^{3}}+{{({{\cos }^{2}}x)}^{3}}dx}{{{\sin }^{2}}x{{\cos }^{2}}x}}\]

\[=\int{\frac{({{\sin }^{2}}x+{{\cos }^{2}}x)\,\,({{\sin }^{4}}x-{{\sin }^{2}}x{{\cos }^{2}}x+{{\cos }^{4}}x)}{{{\sin }^{2}}x{{\cos }^{2}}x}\,\,dx}\]

\[=\int{\frac{{{\sin }^{4}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}\,\,dx}+\int{\frac{{{\cos }^{4}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}}\,\,dx\]

\[-\int{\frac{{{\sin }^{2}}x{{\cos }^{2}}x}{{{\sin }^{2}}x\cdot {{\cos }^{2}}x}\,\,dx}\]

\[=\int{{{\tan }^{2}}}x\,dx+\int{{{\cot }^{2}}x\,dx+-\int{1\,dx}}\]

\[=\int{({{\sec }^{2}}}x-1)\,\,dx+\int{(\cos e{{c}^{2}}x-1)\,\,dx-\int{1\,dx}}\]

\[=\int{{{\sec }^{2}}}\,\,x\,dx+\int{\cos e{{c}^{2}}\,\,x\,dx-3\int{dx}}\]

\[=\tan x-\cot x-3x+C\]