Đáp án: $I=-\cot x+\dfrac{1}{\sin x}+C$
Giải thích các bước giải:
Ta có:
$ I=\displaystyle\int\dfrac{1}{\cos x+1}dx$
$\to I=\displaystyle\int\dfrac{1-\cos x}{(1-\cos x)(1+\cos x)}dx$
$\to I=\displaystyle\int\dfrac{1-\cos x}{1-\cos^2x}dx$
$\to I=\displaystyle\int\dfrac{1-\cos x}{\sin^2x}dx$
$\to I=\displaystyle\int\dfrac{1}{\sin^2x}-\dfrac{\cos x}{\sin^2x}dx$
$\to I=\displaystyle\int\dfrac{1}{\sin^2x}dx-\displaystyle\int\dfrac{\cos x}{\sin^2x}dx$
$\to I=-\cot x-\displaystyle\int\dfrac{1}{\sin^2x}d(\sin x)$
$\to I=-\cot x-(-\dfrac{1}{\sin x})+C$
$\to I=-\cot x+\dfrac{1}{\sin x}+C$
