Question

Encontre o comprimento do arco da função y=ln(cosx) no intervalo 0 a pi/4

Answer 1

Resposta:

Explicação passo-a-passo:

$\int\limits^b_a {\sqrt{1+[f'(x)]^{2}}}\,dx\\\\f'(x)=\frac{1}{cosx}*senx=tgx\\\\1+[f'(x)]^{2}=1+tg^{2}x=sec^{2}x\\\\\int\limits^\frac{\pi }{4} _0 {\sqrt{sec^{2}x}}\,dx=\\\\\int\limits^\frac{\pi }{4} _0 {secx}\,dx=ln|secx+tgx|]\left{ {{\frac{\pi }{4} } \atop {0}}\right.=ln[sec\frac{\pi }{4}+tg\frac{\pi }{4}-(sec0+tg0)]=ln(\sqrt{2}+1-1-0)=ln\sqrt{2}$