$\Rightarrow \int\limits_{0}^{\frac{\pi }{4}}(\frac{1}{cos^2x}-1)^2dx=\int\limits_{0}^{\frac{\pi }{4}}(\frac{1}{cos^4x}-\frac{2}{cos^2x}+1)dx$$= \int\limits_{0}^{\frac{\pi }{4}}\frac{d(tanx)}{cos^2x}-2\int\limits_{0}^{\frac{\pi }{4}}d(tanx)+ \int\limits_{0}^{\frac{\pi }{4}}dx$$ =\int\limits_{0}^{\frac{\pi }{4}}(tan^2x+1)d(tanx) -(2tanx+x)|^{\frac{\pi }{4}}_0$bạn giải tiếp nhé
$\Rightarrow \int\limits_{0}^{\frac{\pi }{4}}(\frac{1}{cos^2x}-1)^2dx=\int\limits_{0}^{\frac{\pi }{4}}(\frac{1}{cos^4x}-\frac{2}{cos^2x}+1)dx$$= \int\limits_{0}^{\frac{\pi }{4}}\frac{d(tanx)}{cos^2x}-2\int\limits_{0}^{\frac{\pi }{4}}d(tanx)+ \int\limits_{0}^{\frac{\pi }{4}}x$$ =\int\limits_{0}^{\frac{\pi }{4}}(tan^2x+1)d(tanx) -(2tanx+x)|^{\frac{\pi }{4}}_0$bạn giải tiếp nhé
$\Rightarrow \int\limits_{0}^{\frac{\pi }{4}}(\frac{1}{cos^2x}-1)^2dx=\int\limits_{0}^{\frac{\pi }{4}}(\frac{1}{cos^4x}-\frac{2}{cos^2x}+1)dx$$= \int\limits_{0}^{\frac{\pi }{4}}\frac{d(tanx)}{cos^2x}-2\int\limits_{0}^{\frac{\pi }{4}}d(tanx)+ \int\limits_{0}^{\frac{\pi }{4}}
dx$$ =\int\limits_{0}^{\frac{\pi }{4}}(tan^2x+1)d(tanx) -(2tanx+x)|^{\frac{\pi }{4}}_0$bạn giải tiếp nhé