Thiếu dx kìa
\(=\int\dfrac{1+x\sin x}{\cos^2x}.dx=\int\dfrac{1}{\cos^2x}.dx+\int\dfrac{x.\sin x}{\cos^2x}.dx\)
\(=\tan x+\int\dfrac{x.\sin x}{\cos^2x}.dx\)
Xet \(\int\dfrac{x.\sin x}{\cos^2x}.dx\)
\(\left\{{}\begin{matrix}u=x\\dv=\dfrac{\sin x}{\cos^2x}.dx\end{matrix}\right.\Rightarrow\int\dfrac{x.\sin x}{\cos^2x}.dx=\dfrac{x}{\cos x}-\int\dfrac{1}{\cos x}.dx\)
\(Co:-\int\dfrac{1}{\cos x}.dx=\int\dfrac{\cos x}{\sin^2x-1}.dx=\int\dfrac{1}{\left(\sin x-1\right)\left(\sin x+1\right)}.d\left(\sin x\right)\)
\(=\dfrac{1}{2}ln\left|\dfrac{\sin x-1}{\sin x+1}\right|\)
\(\Rightarrow\int\dfrac{1+x\sin x}{\cos^2x}.dx=\tan x+\dfrac{x}{\cos x}+\dfrac{1}{2}ln\left|\dfrac{\sin x-1}{\sin x+1}\right|+C\)
