A) 0
B) -1
C) 1
D) \[\pi \]
Correct Answer: A
Solution :
[a] : Let \[I=\int\limits_{{}}^{\pi }{{{e}^{{{\sin }^{2}}x}}{{\cos }^{3}}xdx}\] \[=\int\limits_{0}^{\pi /2}{[{{e}^{{{\sin }^{2}}x}}{{\cos }^{3}}x+{{e}^{{{\sin }^{2}}(\pi -x)}}}{{\cos }^{3}}(\pi -x)]dx\] \[\left[ \because \int\limits_{0}^{2a}{f(x)dx}=\int\limits_{0}^{a}{f(x)}dx+\int\limits_{0}^{a}{f(2a-x)dx} \right]\] \[=\int\limits_{0}^{\pi /2}{\left[ {{e}^{{{\sin }^{2}}x}}{{\cos }^{3}}x-{{e}^{{{\sin }^{2}}x}}{{\cos }^{3}}x \right]}dx=0\]
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