Chủ đề này có 2 trả lời

[rabbit]

$\int_{0}^{\frac{\prod }{4}}\frac{(\left ( sinx \right )^4-\left ( cosx \right )^4)dx}{sinx+cosx+1}$

[Dahitotn94]

$\int_{0}^{\frac{\prod }{4}}\frac{(\left ( sinx \right )^4-\left ( cosx \right )^4)dx}{sinx+cosx+1}$

$I=\int_{0}^{\frac{\pi}{4}}\frac{sin^{2}x-cos^{2}x}{sinx+cosx+1}dx=\int_{0}^{\frac{\pi}{4}}\frac{(sinx+cosx+1)(sinx-cosx)+cosx-sinx}{sinx+cosx+1}dx=\int_{0}^{\frac{\pi}{4}}(sinx-cosx)dx+\int_{0}^{\frac{\pi}{4}}\frac{cosx-sinx}{sinx+cosx+1}dx=J+K$

Tính J thì dễ rồi

$K=\int_{0}^{\frac{\pi}{4}}\frac{d(sinx+cosx)}{sinx+cosx+1}$

ok

[Dahitotn94]

$I=\int_{0}^{1}\frac{1}{x^{4}+1}dx$