Question

Sa se calculeze integrala:
An nevoie de d).
Poza este atașata mai jos

Answer 1

Răspuns

Explicație pas cu pas:

$\int_0^{\frac{\pi}{3}} \frac{tg~x+tg~\frac{\pi}{4}}{9\cos^2x+\sin^2x}dx=\int_0^{\frac{\pi}{3}} \frac{1}{\cos^2 x}\cdot \frac{tg~x+1}{9+\frac{\sin^2x}{\cos^2x}}=\\\int_0^{\frac{\pi}{3}} \frac{1}{\cos^2 x}\cdot \frac{tg~x+1}{9+tg^2x}dx \\tg x =u  => \frac{dx}{\cos^2x}=du\\ Integrala~devine:\\\int_0^{\frac{\pi}{3}} \frac{u+1}{u^2+9} du =\int_0^{\frac{\pi}{3}} \frac{u}{u^2+9} du+\int_0^{\frac{\pi}{3}} \frac{1}{u^2+9} du = \frac{1}{2}\ln(u^2+9) |_0^{\frac{\pi}{3}}+\frac{1}{3} arctg~\frac{u}{3}|_0^{\frac{\pi}{3}} =$

$\frac{1}{2}\ln(\frac{\pi^2}{9}+9)+arctg \frac{\pi}{9}$