$\int\limits^{\frac{\pi}{4}}_0 \dfrac{tan^2x}{cos^2x} \, dx$ 02/11/2021 Bởi Valentina $\int\limits^{\frac{\pi}{4}}_0 \dfrac{tan^2x}{cos^2x} \, dx$
$=> \int\limits^{\tfrac{\pi}{4}}_0 \dfrac{\tan^2x}{\cos^2x} \, dx\\ =\int\limits^{\tfrac{\pi}{4}}_0 \tan^2x\, d(\tan x)\\ =\dfrac{1}{3}\tan^3\Bigg\vert^{\tfrac{\pi}{4}}_0\\ =\dfrac{1}{3}$ Bình luận
$\int\limits^{\tfrac{\pi}{4}}_0 \dfrac{\tan^2x}{\cos^2x} \, dx\\ =\int\limits^{\tfrac{\pi}{4}}_0 \tan^2x\, d(\tan x)\\ =\dfrac{1}{3}\tan^3\Bigg\vert^{\tfrac{\pi}{4}}_0\\ =\dfrac{1}{3}$ Bình luận
$=> \int\limits^{\tfrac{\pi}{4}}_0 \dfrac{\tan^2x}{\cos^2x} \, dx\\ =\int\limits^{\tfrac{\pi}{4}}_0 \tan^2x\, d(\tan x)\\ =\dfrac{1}{3}\tan^3\Bigg\vert^{\tfrac{\pi}{4}}_0\\ =\dfrac{1}{3}$
$\int\limits^{\tfrac{\pi}{4}}_0 \dfrac{\tan^2x}{\cos^2x} \, dx\\ =\int\limits^{\tfrac{\pi}{4}}_0 \tan^2x\, d(\tan x)\\ =\dfrac{1}{3}\tan^3\Bigg\vert^{\tfrac{\pi}{4}}_0\\ =\dfrac{1}{3}$