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 \[u=0\] \[\therefore \]\[l=\log \sqrt{2}[t]_{0}^{\pi /4}+\int\limits_{0}^{\pi /4}{\log \,(\cos u)\,du-\int\limits_{0}^{\pi /4}{\log \,(\cos t)\,dt}}\] \[=\log \sqrt{2}\left[ \frac{\pi }{4}-0 \right]\] \[=\frac{\pi }{8}\log 2\] \[[\because \log {{a}^{b}}=b\log a]\]
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