Definite Integration Question 477
Question: If $ \int_0^{\pi }{xf({{\cos }^{2}}x+{{\tan }^{4}}x)dx} $
$ =k\int_0^{\pi /2}{f({{\cos }^{2}}x+{{\tan }^{4}}x)dx,} $ then the value of $ k $ is
Options:
A) $ \frac{\pi }{2} $
B) $ \pi $
C) $ -\frac{\pi }{2} $
D) None of these
Show Answer
Answer:
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} $
therefore $ 2I=2\pi \int_0^{\pi /2}{f({{\cos }^{2}}x+{{\tan }^{4}}x)dx} $
therefore $ I=\pi \int_0^{\pi /2}{f({{\cos }^{2}}x+{{\tan }^{4}}x)dx} $
On comparing with given integral, we get $ k=\pi $ .
BETA
