Find the exact length of the curve. y = ln(sec(x)). 0 ≤ x ≤ π/4

Determine the length of the curve y = ln(sec(x)) between 0 ? x ? ?/4 .

Answer

we have y=ln(sec(x)) frac{dy}{dx}=frac{d}{dx}ln(sec(x)) frac{dy}{dx}=frac{1}{sec(x)}frac{d}{dx}(sec(x)) frac{dy}{dx}=frac{1}{sec(x)}(sec(x)tan(x)) frac{dy}{dx}=tan(x) the length of the curve is, L=int_{a}^{b}sqrt{1+left (frac{dy}{dx} right )^2} dx now put a = 0, b = pi/4 and value of dy/dx in above equation, L=int_{0}^{frac{pi}{4}}sqrt{1+tan^2(x)} dx L=int_{0}^{frac{pi}{4}}sqrt{sec^2(x)} dx L=int_{0}^{frac{pi}{4}}sec(x) dx L=left [ lnleft |tan(x)+sec(x) right | right ]_{0}^{frac{pi}{4}} L=left [ lnleft |tanleft ( frac{pi}{4} right )+secleft ( frac{pi}{4} right ) right | right ]-left [ ln left |tan(0)+sec(0) right | right ] L=left [ lnleft |1+sqrt{2} right | right ]-left [ ln left |0+1 right | right ] L=left [ lnleft |1+sqrt{2} right | right ]-left [ ln left |1 right | right ] L=left [ lnleft |1+sqrt{2} right | right ]-0 L=lnleft |1+sqrt{2} right | L=0.88137

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