Tìm nguyên hàm của hàm số sau: ∫tan(x+π3)cot(x+π6)dx

Question

\(\int \tan \left(x+\frac{\pi }{3} \right)\cot \left(x+\frac{\pi }{6} \right) {\rm d}x.\)

Answer 1

Ta có:
\(\[\begin{array}{l} {\tan \left(x+\frac{\pi }{3} \right)\cot \left(x+\frac{\pi }{6} \right)=\frac{\sin \left(x+\frac{\pi }{3} \right)\cos \left(x+\frac{\pi }{6} \right)}{\cos \left(x+\frac{\pi }{3} \right)\sin \left(x+\frac{\pi }{6} \right)} } \\ {=\frac{\sin \left(x+\frac{\pi }{3} \right)\cos \left(x+\frac{\pi }{6} \right)-\cos \left(x+\frac{\pi }{3} \right)\sin \left(x+\frac{\pi }{6} \right)}{\cos \left(x+\frac{\pi }{3} \right)\sin \left(x+\frac{\pi }{6} \right)} +1} \\ {=\frac{\sin \left[\left(x+\frac{\pi }{3} \right)-\left(x+\frac{\pi }{6} \right)\right]}{\cos \left(x+\frac{\pi }{3} \right)\sin \left(x+\frac{\pi }{6} \right)} +1=\frac{1}{2} .\frac{1}{\cos \left(x+\frac{\pi }{3} \right)\sin \left(x+\frac{\pi }{6} \right)} +1} \end{array}\] \)
Suy ra:\( I=\frac{1}{2} \int \frac{1}{\cos \left(x+\frac{\pi }{3} \right)\sin \left(x+\frac{\pi }{6} \right)} {\rm d}x +\int {\rm d}x =\frac{1}{2} I_{1} +x+C. \)

 Tính \( I_{1} =\int \frac{1}{\cos \left(x+\frac{\pi }{3} \right)\sin \left(x+\frac{\pi }{6} \right)} {\rm d}x \)

 Ta có \(1=\frac{\cos \frac{\pi }{6} }{\cos \frac{\pi }{6} } =\frac{\cos \left[\left(x+\frac{\pi }{3} \right)-\left(x+\frac{\pi }{6} \right)\right]}{\frac{\sqrt{3} }{2} } =\frac{2\sqrt{3} }{3} \left[\cos \left(x+\frac{\pi }{3} \right)\cos \left(x+\frac{\pi }{6} \right)+\sin \left(x+\frac{\pi }{3} \right)\sin \left(x+\frac{\pi }{6} \right)\right]\)
\(I_{1} =\int \frac{\frac{2\sqrt{3} }{3} \left[\cos \left(x+\frac{\pi }{3} \right)\cos \left(x+\frac{\pi }{6} \right)+\sin \left(x+\frac{\pi }{3} \right)\sin \left(x+\frac{\pi }{6} \right)\right]}{\cos \left(x+\frac{\pi }{3} \right)\sin \left(x+\frac{\pi }{6} \right)} {\rm d}x \)
\(=\frac{2\sqrt{3} }{3} \left[\int \frac{\cos \left(x+\frac{\pi }{6} \right)}{\sin \left(x+\frac{\pi }{6} \right)} {\rm d}x +\int \frac{\sin \left(x+\frac{\pi }{3} \right)}{\cos \left(x+\frac{\pi }{3} \right)} {\rm d}x \right]\)
\(I_{1} =\frac{2\sqrt{3} }{3} \left[\int \frac{{\rm d}\left(\sin \left(x+\frac{\pi }{6} \right)\right)}{\sin \left(x+\frac{\pi }{6} \right)} {\rm d}x -\int \frac{{\rm d}\left(\cos \left(x+\frac{\pi }{3} \right)\right)}{\cos \left(x+\frac{\pi }{3} \right)} {\rm d}x \right]=\frac{2\sqrt{3} }{3} \ln \left|\frac{\sin \left(x+\frac{\pi }{6} \right)}{\cos \left(x+\frac{\pi }{3} \right)} \right|+C_{1} .\)
Vậy \( I=\frac{\sqrt{3} }{3} \ln \left|\frac{\sin \left(x+\frac{\pi }{6} \right)}{\cos \left(x+\frac{\pi }{3} \right)} \right|+x+C.\)