29Both the statements looks correct...i think it shud be A
statement 1: 0∫Π1/1+5cosx=Π/2
statement 2:0∫af(x)dx=0∫af(a-x)dx
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4 Answers
29Using the property \int_{a}^{b}{f(x)dx} = \int_{a}^{b}{f(a+b -x)dx}
we have
I = \int_{0}^{\pi}{\frac{dx}{1+5^{cosx}}} = \int_{0}^{\pi}{\frac{dx}{1+5^{-cosx}}} = \int_{0}^{\pi}{\frac{5^{cosx}dx}{1+5^{cosx}}}
so now 2I = \int_{0}^{\pi}{\frac{1+5^{cosx}}{1+5^{cosx}}}dx = \int_{0}^{\pi}{dx} = \pi
so I = π/2
6oh i got till the last step but didnt wrote I instead of 2I.......thx.....