Mình tính nguyên hàm, bạn tự thế cận nhé!Tính $I= \int {\frac{{\cos (x - \frac{\pi }{4})}}{{4 - 3\cos x}}} dx$Ta có: $\frac{{\cos (x - \frac{\pi }{4})}}{{4 - 3\cos x}} = \frac{{\sqrt 2 }}{2}.\frac{{\sin x + \cos x}}{{4 - 3\cos x}} $ $= \frac{{\sqrt 2 }}{2}.\frac{{\sin x}}{{4 - 3\cos x}} + \frac{{\sqrt 2 }}{2}\left( { - \frac{1}{3} + \frac{4}{3}.\frac{1}{{4 - 3\cos x}}} \right)$$ \Rightarrow I = \frac{{\sqrt 2 }}{2}\int {\frac{{\sin x}}{{4 - 3\cos x}}} dx - \frac{{\sqrt 2 }}{6}\int {dx} + \frac{{2\sqrt 2 }}{3}\int {\frac{1}{{4 - 3\cos x}}} dx$* Tính $A = \int {\frac{{\sin x}}{{4 - 3\cos x}}} dx$: đặt $t = 4 - 3\cos x$* Tính $B = \int {\frac{1}{{4 - 3\cos x}}} dx$$4 - 3\cos x = 4({\sin ^2}\frac{x}{2} + {\cos ^2}\frac{x}{2}) - 3({\cos ^2}\frac{x}{2} - {\sin ^2}\frac{x}{2}) = 7{\sin ^2}\frac{x}{2} + {\cos ^2}\frac{x}{2} = {\cos ^2}\frac{x}{2}(7{\tan ^2}\frac{x}{2} + 1)$$\Rightarrow B = \int {\frac{1}{{{{\cos }^2}\frac{x}{2}(7{{\tan }^2}\frac{x}{2} + 1)}}} dx$Đặt $t = \tan \frac{x}{2} \Leftrightarrow 2dt = \frac{{dx}}{{{{\cos }^2}\frac{x}{2}}}$$\Rightarrow B = \int {\frac{{2dt}}{{7{t^2} + 1}}} $Tới đây đặt tiếp $t = \frac{{\tan u}}{{\sqrt 7 }}$ là xong.
Mình tính nguyên hàm, bạn tự thế cận nhé!Tính $I= \int {\frac{{\cos (x - \frac{\pi }{4})}}{{4 - 3\cos x}}} dx$Ta có: $\frac{{\cos (x - \frac{\pi }{4})}}{{4 - 3\cos x}} = \frac{{\sqrt 2 }}{2}.\frac{{\sin x + \cos x}}{{4 - 3\cos x}} $ $= \frac{{\sqrt 2 }}{2}.\frac{{\sin x}}{{4 - 3\cos x}} + \frac{{\sqrt 2 }}{2}\left( { - \frac{1}{3} + \frac{4}{3}.\frac{1}{{4 - 3\cos x}}} \right)$$ \Rightarrow I = \frac{{\sqrt 2 }}{2}\int {\frac{{\sin x}}{{4 - 3\cos x}}} dx - \frac{{\sqrt 2 }}{6}\int {dx} + \frac{{2\sqrt 2 }}{3}\int {\frac{1}{{4 - 3\cos x}}} dx$* Tính $*A = \int {\frac{{\sin x}}{{4 - 3\cos x}}} dx$: đặt $t = 4 - 3\cos x$* Tính $B = \int {\frac{1}{{4 - 3\cos x}}} dx$$4 - 3\cos x = 4({\sin ^2}\frac{x}{2} + {\cos ^2}\frac{x}{2}) - 3({\cos ^2}\frac{x}{2} - {\sin ^2}\frac{x}{2}) = 7{\sin ^2}\frac{x}{2} + {\cos ^2}\frac{x}{2} = {\cos ^2}\frac{x}{2}(7{\tan ^2}\frac{x}{2} + 1)$$\Rightarrow B = \int {\frac{1}{{{{\cos }^2}\frac{x}{2}(7{{\tan }^2}\frac{x}{2} + 1)}}} dx$Đặt $t = \tan \frac{x}{2} \Leftrightarrow 2dt = \frac{{dx}}{{{{\cos }^2}\frac{x}{2}}}$$\Rightarrow B = \int {\frac{{2dt}}{{7{t^2} + 1}}} $Tới đây đặt tiếp $t = \frac{{\tan u}}{{\sqrt 7 }}$ là xong.
Mình tính nguyên hàm, bạn tự thế cận nhé!Tính $I= \int {\frac{{\cos (x - \frac{\pi }{4})}}{{4 - 3\cos x}}} dx$Ta có: $\frac{{\cos (x - \frac{\pi }{4})}}{{4 - 3\cos x}} = \frac{{\sqrt 2 }}{2}.\frac{{\sin x + \cos x}}{{4 - 3\cos x}} $ $= \frac{{\sqrt 2 }}{2}.\frac{{\sin x}}{{4 - 3\cos x}} + \frac{{\sqrt 2 }}{2}\left( { - \frac{1}{3} + \frac{4}{3}.\frac{1}{{4 - 3\cos x}}} \right)$$ \Rightarrow I = \frac{{\sqrt 2 }}{2}\int {\frac{{\sin x}}{{4 - 3\cos x}}} dx - \frac{{\sqrt 2 }}{6}\int {dx} + \frac{{2\sqrt 2 }}{3}\int {\frac{1}{{4 - 3\cos x}}} dx$* Tính $A = \int {\frac{{\sin x}}{{4 - 3\cos x}}} dx$: đặt $t = 4 - 3\cos x$* Tính $B = \int {\frac{1}{{4 - 3\cos x}}} dx$$4 - 3\cos x = 4({\sin ^2}\frac{x}{2} + {\cos ^2}\frac{x}{2}) - 3({\cos ^2}\frac{x}{2} - {\sin ^2}\frac{x}{2}) = 7{\sin ^2}\frac{x}{2} + {\cos ^2}\frac{x}{2} = {\cos ^2}\frac{x}{2}(7{\tan ^2}\frac{x}{2} + 1)$$\Rightarrow B = \int {\frac{1}{{{{\cos }^2}\frac{x}{2}(7{{\tan }^2}\frac{x}{2} + 1)}}} dx$Đặt $t = \tan \frac{x}{2} \Leftrightarrow 2dt = \frac{{dx}}{{{{\cos }^2}\frac{x}{2}}}$$\Rightarrow B = \int {\frac{{2dt}}{{7{t^2} + 1}}} $Tới đây đặt tiếp $t = \frac{{\tan u}}{{\sqrt 7 }}$ là xong.