Let [x] denote the largest integer not exceeding x and \x\=x-[x] . Then ∫_0^2012 frace^cos (π\x\)e^cos (π\x\)+e^-cos (π\x\) d x is equal to

Question

Let [x] denote the largest integer not exceeding x and \x\=x-[x] . Then ∫_0^2012 frace^cos (π\x\)e^cos (π\x\)+e^-cos (π\x\) d x is equal to -

MARKS Content Team · Accepted Answer

Correct answer: (B) 1006. I=2012 ∫_0^1 frace^cos π xe^cos π x+e^-cos π x d x using king property I=2012 ∫_0^1 frace^-cos π xe^-cos π x+e^cos π x d x ⇒ 2 I=2012 ⇒ I=1006

Let $[x]$ denote the largest integer not exceeding $x$ and $\{x\}=x-[x]$. Then $\int_{0}^{2012} \frac{e^{\cos (\pi\{x\})}}{e^{\cos (\pi\{x\})}+e^{-\cos (\pi\{x\})}} d x$ is equal to -

Solution

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