[TEX]I=\int_0^{\frac{\pi}{2}}\frac{sinx}{(sinx+\sqrt3cosx)^3}dx[/TEX]
[TEX]1/\ Cach\ tra\ bai\ kha\ dai\ dong \ nhung\ lai\ thuong\ hay\ su\ dung:[/TEX]
[TEX]I=\int_0^{\frac{\pi}{4}}\frac{sinx}{(sinx+\sqrt3cosx)^3}dx+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{sinx}{(sinx+\sqrt3cosx)^3}dx=I_1+I_2[/TEX]
[TEX]\left{I_1\ :\ t=tgx \\I_2\ :\ t=cotgx[/TEX]
[TEX]2/\ Cach\ dat\ t=tgx\ roi\ su\ dung\ gioi\ han\ :\ khong\ thich\ hop\ cho\ chuong\ trinh\ pho\ thong[/TEX]
[TEX]3/\ Cach\ tu\ duy\ khong\ quan\ tam\ den\ can\ ,\ tinh\ duoc\ luon\ ca\ nguyen\ ham\ :[/TEX]
[TEX]\sqrt3I=\int_0^{\frac{\pi}{2}}\frac{\sqrt3sinx-cosx+cosx}{(sinx+\sqrt3cosx)^3}dx=-\int_0^{\frac{ \pi}{2}}\frac{1}{(sinx+\sqrt3cosx)^3}d(sinx+\sqrt3cosx)+\int_0^{\frac{\pi}{2}}\frac{cosx}{(sinx+ \sqrt3 cosx)^3}dx[/TEX]
[TEX]=\frac{1}{2}\frac{1}{(sinx+\sqrt3 cosx)^2}\|_0^{\frac{\pi}{2}}+I_1=\frac{1}{3}+I_1[/TEX][TEX]\Rightarrow{\sqrt3I_1=3I-\frac{\sqrt3}{3}[/TEX]
[TEX]I+3I-\frac{\sqrt3}{3}=\int_0^{\frac{\pi}{2}}\frac{sinx+\sqrt3cosx}{(sinx+\sqrt3cosx)^3}dx=\int_0^{ \frac{\pi}{2}}\frac{1}{4cos^2(x-\frac{\pi}{6})}dx=\frac{1}{4}tg(x-\frac{\pi}{6})\|_0^{\frac{\pi}{2}}=\frac{\sqrt3}{3}\Leftrightarrow{I=\frac{\sqrt3}{6}[/TEX]
[TEX]*\ Nhan\ xet\ :\ trong\ cac\ bai\ toan\ co\ ham\ luong\ giac\ o\ mau\ so\ thi\ ta \ nen\ tach\ sao[/TEX][TEX]\ cho\ tu\so\ chua\ bieu\ thuc\ mau\ so\ va\ dao\ ham\ cua\ mau\ de\ dua\ ve\ cac\ nguyen\ ham\ don\ gian\ ,\ dong\ thoi\ lien\ ket\ cac[/TEX]
[TEX]\ tich\ phan\ co\ dang\ tuong\ tu\ nhau\ lai\ de\ giai\ phuong\ trinh\ nhu\ bai\ toan\ tren[/TEX]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
