use the property a∫b f(x) = a∫b f(a+b-x)
I = \int_{0}^{\pi }{\frac{dx}{1+5^{cosx}}} = \int_{0}^{\pi }{\frac{dx}{1+5^{-cosx}}}
2I = \int_{0}^{\pi }dx \Rightarrow I = \frac{\pi }{2}
use the property a∫b f(x) = a∫b f(a+b-x)
I = \int_{0}^{\pi }{\frac{dx}{1+5^{cosx}}} = \int_{0}^{\pi }{\frac{dx}{1+5^{-cosx}}}
2I = \int_{0}^{\pi }dx \Rightarrow I = \frac{\pi }{2}
Since f(x) = f(-x), this is an even function.
\int_{0}^{\pi }{\frac{1}{1+5^{cosx}}} = 2\int_{0}^{\frac{\pi}{2} }{\frac{1}{1+5^{cosx}}}
2\int_{0}^{\frac{\pi}{2} }{\frac{1}{1+5^{cosx}}} = 2\int_{0}^{\frac{\pi}{2} }{\frac{1}{1+5^{cos(\frac{\pi}{2} - x)}}} = 2\int_{0}^{\frac{\pi}{2} }{\frac{1}{1+5^{sinx}}}
Does this make a difference? lola
haha..govind got it :P
no pritish i thnk it's to approach solution by ur method.
Applying govind's method is the way i wuld have gone for .....
hey thanx that was easy actually i did the same thing but screwed up the adittion LOL ...