Tính nguyên hàm: \(\int {\frac{{{{\cos }^3}x}}{{\sin x + \cos x}}dx} \)

Đáp án:

$\dfrac{x}{2} + \dfrac{\cos2x + \sin2x}{8} + \dfrac{1}{4}\ln|\sin x + \cos x| + C$ Với $C$ là hằng số

Giải thích các bước giải:

$\begin{array}{l}Đặt\,\,I = \displaystyle\int\dfrac{\cos^3x}{\sin x + \cos x}dx\\ J =\displaystyle\int\dfrac{\sin^3x}{\sin x + \cos x}dx \\ +) \quad I + J = \displaystyle\int\dfrac{\cos^3x + \sin^3x}{\sin x + cos x}dx\\ =\displaystyle\int(\sin^2x + \cos^2x - \sin x\cos x)dx\\ = \displaystyle\int\left(1 - \dfrac{1}{2}\sin2x\right)dx\\ = x + \dfrac{1}{4}\cos2x + C_1\qquad (1)\\ +) \quad I-J = \displaystyle\int\dfrac{\cos^3x - \sin^3x}{\sin x + \cos x}dx\\ = \displaystyle\int\dfrac{(\cos x - \sin x)(1+ \cos x\sin x)}{\sin x + \cos x}dx\\ Đặt\,\,u = \sin x + \cos x \quad \Rightarrow du = (\cos x - \sin x)dx\\ \Rightarrow u^2 = 1 + 2\sin x\cos x\\ \Rightarrow \dfrac{u^2 -1}{2} = \sin x\cos x\\ \Rightarrow I - J = \displaystyle\int\dfrac{1 + \dfrac{u^2 -1}{2}}{u}du\\ = \displaystyle\int\dfrac{u^2 + 1}{2u}du\\ = \dfrac{1}{2}\displaystyle\int\left(u + \dfrac{1}{u}\right)du\\ = \dfrac{u^2}{4} + \dfrac{1}{2}\ln|u| + C_2\\ = \dfrac{(\sin x + \cos x)^2}{4} + \dfrac{1}{2}\ln|\sin x + \cos x| + C_2\qquad (2)\\ (1) + (2) = 2I = x + \dfrac{1}{4}\cos2x + \dfrac{1 + \sin2x}{4} + \dfrac{1}{2}\ln|\sin x + \cos x| + C_3\\ \Leftrightarrow I = \dfrac{x}{2} + \dfrac{\cos2x + \sin2x}{8} + \dfrac{1}{4}\ln|\sin x + \cos x| + C\end{array}$