Tach ra thanh $I=\int\limits_{}^{}\frac{21sin^{3}x}{cos^{4}x}dx+\int\limits_{}^{}\frac{8}{cos^{4}x}dx=I_{1}+I_{2}$
$tinh I_{1}. dat cosx=t =>dt=-sinxdx$
$I_{1}=21\int\limits_{}^{}\frac{(t^{2}-1)dt}{t^{4}}=21\int\limits_{}^{}t^{-2}dt-21\int\limits_{}^{}t^{-4}dt$ ( cai nay co ban roi)
$Tinh I_{2}.$ dat $dat tanx=t =>dt=(tan^{2}x+1)dx$
$I_{2}=8\int\limits_{}^{}\frac{1}{cos^{2}x}.\frac{1}{cos^{2}x}dx=8\int\limits_{}^{}(tan^{2}x+1)^{2}dx=\int\limits_{}^{}(t^{2}+1)dt$
cai nay cung qua de roi
P/S: can cau khong danh vao nen loi giai buon cuoi. Thay can , doi can tu lam nhe ban.