Tính tích phân

[muahoatuyet]

\[\int \frac{cos2x dx}{1+sinxcosx}\]
\[\int \frac{dx}{1+2sinx}\]

 

[NguoiDien]

Đặt \[t=2+\sin 2x\Rightarrow dt=2\cos 2xdx\]

khi đó

\[\frac{\cos 2xdx}{1+\sin x\cos x}=\frac{\cos 2xdx}{1+\frac{\sin 2x}{2}}=\frac{2\cos 2xdx}{2+\sin 2x}=\frac{dt}{t}\]

 

[NguoiDien]

Đặt \[t=\tan \frac{x}{2}\Rightarrow dt=\frac{1}{2}(1+\tan ^2\frac{x}{2})dx\] hay \[dx=\frac{2dt}{1+t^2}\]

Khi đó:

\[\frac{dx}{1+2\sin x}=\frac{\frac{2dt}{1+t^2}}{1+\frac{2t}{1+t^2}}= \frac{2dt}{1+t^2+2t}=\frac{2dt}{(1+t)^2}\]