|
|
Trước hết ta có
$1=(\sin^2 x +\cos^2 x)^2=\sin^4 x+\cos^4 x +2\sin^2 x \cos^2 x=\sin^4 x+\cos^4 x +2\sin^2 x \cos^2 x(\sin^2 x +\cos^2 x)$
$=\sin^4 x(1+2\cos^2 x)+\cos^4 x (1+2\sin^2 x)=\sin^4 x(1+2\cos^2 x)+\cos^4 x (\cos^2 x+3\sin^2 x)$
$=\cos^6x+\sin^4 x+2\sin^4 \cos^2 x+3\sin^2 x\cos^4 x$
Suy ra
$\frac{1}{\sin^3x\cos^5x}=\frac{\cos^6x+\sin^4 x+2\sin^4 \cos^2 x+3\sin^2 x\cos^4 x}{\sin^3x\cos^5x}=\frac{\cos x}{\sin^3 x}+\frac{\sin x}{\cos^5 x}+\frac{2\sin x}{\cos^3 x}+\frac{3}{\sin x \cos x}$
$\Rightarrow \int\limits\frac{1}{\sin^3x\cos^5x}dx=\int\limits\frac{\cos x}{\sin^3 x}dx+\int\limits\frac{\sin x}{\cos^5 x}dx+\int\limits\frac{2\sin x}{\cos^3 x}dx+\int\limits\frac{3}{\sin x \cos x}dx$
$\Rightarrow \int\limits\frac{1}{\sin^3x\cos^5x}dx=\int\limits\frac{d(\sin^2
x)}{2\sin^4 x}-\int\limits\frac{d(\cos^4 x)}{4\cos^8
x}-\int\limits\frac{d(\cos^2 x)}{\cos^4
x}+3\int\limits\frac{d(\tan x)}{\tan x}$
Vậy
$\boxed{\displaystyle{ \int\limits\frac{1}{\sin^3x\cos^5x}dx=-\frac{1}{2\sin^2x}+\frac{1}{4\cos^4x}+\frac{1}{\cos^2x}+3\ln|\tan x|+C}}$
|