A) 0
B) 1
C) \[{{e}^{1/2}}\]
D) \[2{{e}^{1/2}}\]
Correct Answer: A
Solution :
\[I=\int_{-1/2}^{1/2}{(\cos x)\left[ \log \left( \frac{1-x}{1+x} \right) \right]dx}\] ?..(i) \ \[I=\int_{-1/2}^{1/2}{\cos (-x)\left[ \log \left( \frac{1+x}{1-x} \right) \right]}\,dx\] Þ \[I=-\int_{-1/2}^{1/2}{\cos x\left[ \log \left( \frac{1-x}{1+x} \right) \right]}\,dx\] ?..(ii) Adding (i) and (ii), we get \[2I=\int_{\,-1/2}^{\,1/2}{\cos \,x\,\left[ \log \,\left( \frac{1-x}{1+x} \right) \right]}\,dx-\int_{\,-1/2}^{\,1/2}{\cos \,x\,\left[ \log \,\left( \frac{1-x}{1+x} \right) \right]\,\,dx}\] or \[2I=0\]or I = 0.
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