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Evaluate:\(\int\limits_0^{\pi/2}\frac{dx}{1+cos^2x}\) |
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Answer» \(\int\limits_0^{\pi/2}\frac{dx}{1+cos^2x}=\int\limits_{0}^{\pi/2}\frac{sec^2xdx}{sec^2x+1}\) = \(\int\limits_0^{\pi/2}\frac{sec^2xdx}{2+tan^2x}\) Let tan x = t ⇒ sec2 x dx = dt x = 0 then t = 0 & when x = \(\frac{\pi}2\) then t = \(\infty\) = \(\int\limits_0^{\infty}\frac{dt}{2+t^2}\) = \(\frac1{\sqrt2}[tan^{-1}\frac{x}{\sqrt2}]_0^{\infty}\) = \(\frac1{\sqrt2}(tan^-1\infty - tan^{-1}0)\) = \(\frac1{\sqrt2}(\frac{\pi}2-0)=\frac{\pi}{2\pi}\) |
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