A) \[\frac{\pi }{2}\]
B) \[\pi \]
C) \[-\frac{\pi }{2}\]
D) None of these
Correct Answer: B
Solution :
\[\int_{0}^{\pi }{xf}({{\cos }^{2}}x+{{\tan }^{4}}x)dx=k\int_{0}^{^{\pi /2}}{f({{\cos }^{2}}x+{{\tan }^{4}}x)dx}\] By the property of definite integral \[I=\int_{0}^{\pi }{x}f({{\cos }^{2}}x+{{\tan }^{4}}x)dx\] ?..(i) \[=\int_{0}^{\pi }{(\pi -x)f({{\cos }^{2}}x+{{\tan }^{4}}x)dx}\] ?..(ii) Adding (i) and (ii), we have \[2I=\pi \int_{0}^{\pi }{f({{\cos }^{2}}x+{{\tan }^{4}}x)dx}\] Þ \[2I=2\pi \int_{0}^{\pi /2}{f\,({{\cos }^{2}}x+{{\tan }^{4}}x)dx}\] Þ \[I=\pi \int_{0}^{\pi /2}{f({{\cos }^{2}}x+{{\tan }^{4}}x)dx}\] On comparing with given integral, we get \[k=\pi \].
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