question_answer

Evaluate \[\int_{3}^{4}{\frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}}dx.}\]

OR

Evaluate \[\int_{0}^{1}{\frac{\log (1+x)}{1+{{x}^{2}}}\,dx.}\]

Answer:

Let \[l=\int\limits_{3}^{4}{\frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}}\,\,dx}\]                                 ?(i)

Using \[\int\limits_{a}^{b}{f(x)dx=}\int\limits_{a}^{b}{f(a+b-x)dx}\] in Eq. (i), we get

\[l=\int\limits_{3}^{4}{\frac{\sqrt{3+4-x}}{\sqrt{3+4-x}+\sqrt{7-(3+4-x)}}dx}\]

\[=\int\limits_{3}^{4}{\frac{\sqrt{7-x}}{\sqrt{7-x}+\sqrt{x}}}\]

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

\[2l=\int\limits_{3}^{4}{\frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}}dx+\int\limits_{3}^{4}{\frac{\sqrt{7-x}}{\sqrt{7-x}+\sqrt{x}}dx}}\]

\[=\int\limits_{3}^{4}{\frac{\sqrt{x}+\sqrt{7-x}}{\sqrt{x}+\sqrt{7-x}}dx+\int\limits_{3}^{4}{dx=[x]_{3}^{4}}}\]

\[\Rightarrow \]   \[2l=[4-3]=1\]

\[\therefore \]      \[l=\frac{1}{2}\]

OR

Let \[l=\int\limits_{0}^{1}{\frac{\log \,(1+x)}{1+{{x}^{2}}}\,\,dx}\]

Put \[t={{\tan }^{-1}}x\Rightarrow dt=\frac{1}{1+{{x}^{2}}}dx\]

\[\therefore \]Also, when \[x=0,\] then \[t=0\]

and when \[x=1,\] then \[t=\frac{\pi }{4}\]

\[\therefore \]      \[l=\int\limits_{0}^{\pi /4}{\log \,(tant\,+1)dt=\int\limits_{0}^{\pi /4}{\log \left( \frac{\sin t}{\cos t}+1 \right)dt}}\]

\[=\int\limits_{0}^{\pi /4}{\log \left( \frac{\sin t+\cos t}{\cos t} \right)}\,\,dt\]

\[=\int\limits_{0}^{\pi /4}{[log\,(sint+cost)-\log \,(cost)]dt}\]

Here, \[\sin t+\cos t=\sqrt{2}\left( \frac{1}{\sqrt{2}}\sin t+\frac{1}{\sqrt{2}}\cos t \right)\]

\[=\sqrt{2}\left[ \sin \frac{\pi }{4}\cdot \sin t+\cos \frac{\pi }{4}\cdot \cos t \right]\]

\[=\sqrt{2}\cos \left( \frac{\pi }{4}-t \right)\]

Now,     \[l=\int\limits_{0}^{\pi /4}{\left[ \log \left[ \sqrt{2}\cos \left( \frac{\pi }{4}-t \right) \right]-\log (cost) \right]dt}\]

\[=\int\limits_{0}^{\pi /4}{\log \sqrt{2}dt+\int\limits_{0}^{\pi /4}{\log \left( \cos \left( \frac{\pi }{4}-t \right) \right)dt}}\]

\[-\int\limits_{0}^{\pi /4}{\log \,(cos\,t)dt}\]

Again, put \[u=\frac{\pi }{4}-t\Rightarrow du=-dt\]

Also, when \[t=0,\]then \[u=\frac{\pi }{4}\]and when \[t=\frac{\pi }{4}\]then \[u=0\]

\[\therefore \]\[l=\log \sqrt{2}[t]_{0}^{\pi /4}+\int\limits_{0}^{\pi /4}{\log \,(\cos u)\,du-\int\limits_{0}^{\pi /4}{\log \,(\cos t)\,dt}}\]

\[=\log \sqrt{2}\left[ \frac{\pi }{4}-0 \right]\]

\[=\frac{\pi }{8}\log 2\]             \[[\because \log {{a}^{b}}=b\log a]\]