Đáp án:
$\int\dfrac{dx}{\cos x}=\dfrac{1}{2}\left[{-\ln(1-\sin x)+\ln(1+\sin x)}\right]$
Giải thích các bước giải:
Đặt $A=\int\dfrac{dx}{\cos x}=\int\dfrac{\cos x}{{\cos}^2x}dx=\int\dfrac{\cos x}{(1-{\sin }^2x)}dx$
Đặt $\sin x=t$ (-1<t<1) $\Rightarrow \cos xdx=dt$
$A=\int\dfrac{dt}{1-t^2}=\int\dfrac{dt}{(1-t)(1+t)}$
$=\dfrac{1}{2}\int\left({\dfrac{1}{1-t}+\dfrac{1}{1+t}}\right)dt=\dfrac{1}{2}\left({-\ln|1-t|}+\ln|1+t|\right)$
$=\dfrac{1}{2}\left({-\ln(1-t)}+\ln(1+t)\right)$ (do -1<t<1)
Vậy $\int\dfrac{dx}{\cos x}=\dfrac{1}{2}\left[{-\ln(1-\sin x)+\ln(1+\sin x)}\right]$
