|
|
b.
Ta có:
$I=\int\frac{dx}{2-\cos^2x}=\int\frac{dx}{2\sin^2x+\cos^2x}=\int\frac{d(\tan x)}{2\tan^2x+1}$
Đặt $\sqrt2\tan x=t$
Ta có: $I=\frac{1}{\sqrt2}\int\frac{dt}{t^2+1}$
Đặt $t=\tan z\Rightarrow dt=(\tan^2z+1)dz$
Suy ra:
$I=\frac{1}{\sqrt2}\int\frac{(\tan^2z+1)dz}{\tan^2z+1}$
$=\frac{1}{\sqrt2}\int dz$
$=\frac{z}{\sqrt2}+C$
$=\frac{\arctan t}{\sqrt2}+C$
$=\frac{\arctan(\sqrt2\tan x)}{\sqrt2}+C$
|