Tìm nguyên hàm của hàm số sau: ∫dx/3√sinx+cosx

Question

\(\int \frac{{\rm d}x}{\sqrt{3} \sin x+\cos x}\)

Answer 1

Ta có:

\(I=\int \frac{{\rm d}x}{\sqrt{3} \sin x+\cos x} =\int \frac{{\rm d}x}{2\left(\frac{\sqrt{3} }{2} \sin x+\frac{1}{2} \cos x\right)}\)

 \(=\frac{1}{2} \int \frac{{\rm d}x}{\sin \left(x+\frac{\pi }{6} \right)}\) \(=\frac{1}{2} \int \frac{\sin \left(x+\frac{\pi }{6} \right)}{\sin ^{2} \left(x+\frac{\pi }{6} \right)} {\rm d}x\)
\(=\frac{1}{2} \int \frac{\sin \left(x+\frac{\pi }{6} \right)}{1-\cos ^{2} \left(x+\frac{\pi }{6} \right)} {\rm d}x .\)

Đặt   

\(t=\cos \left(x+\frac{\pi }{6} \right)\Rightarrow {\rm d}t=-\sin \left(x+\frac{\pi }{6} \right){\rm d}x.\)

 Suy ra:

\(I=-\frac{1}{2} \int \frac{{\rm d}t}{1-t^{2} } =-\frac{1}{2} \int \frac{{\rm d}t}{\left(1-t\right)\left(1+t\right)} =-\frac{1}{4} \int \left(\frac{1}{1+t} +\frac{1}{1-t} \right){\rm d}t =-\frac{1}{4} \left(\ln \left|1+t\right|-\ln \left|1-t\right|\right)+C\)
\(=\frac{1}{4} \left(\ln \left|1-t\right|-\ln \left|1+t\right|\right)+C=\frac{1}{4} \ln \left|\frac{1-t}{1+t} \right|+C.\)
Vậy \(I=\frac{1}{4} \ln \left|\frac{1-\cos \left(x+\frac{\pi }{6} \right)}{1+\cos \left(x+\frac{\pi }{6} \right)} \right|+C.\)