Question

Valoarea integralei din imagine

Answer 1

Calculam intai integrala:

$\int\frac{tgx\cdot (1+\frac{sin^2x}{cos^2x)} }{1+tg^4x} \ dx\\\\\int\frac{tgx\cdot\frac{1}{cos^2x} }{1^2+(tg^2x)^2} \ dx \\\\\int\frac{tgx\cdot tg'x}{1^2+(tg^2x)^2}\ dx\\\\ stim\ ca\ (tg^2x)'=2tgx\cdot tg'x\\\\\frac{1}{2} \int \frac{2tgx\cdot tg'x}{1^2+(tg^2x)^2}\ dx=\frac{1}{2}\cdot arctg(tg^2x) +C\\\\ \frac{1}{2}\cdot arctg(tg^2x) |_0^{\frac{\pi}{3}} =\frac{1}{2}arctg3- \frac{1}{2}arctg0= \frac{1}{2}arctg3$

O alta integrala gasesti aici: https://brainly.ro/tema/3171679

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