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[25 minutes]

Tìm nguyên hàm $I=\int \frac{\sin x \cos 2x}{\cos 5x}dx$

[chinhanh9]

Tìm nguyên hàm $I=\int \frac{\sin x \cos 2x}{\cos 5x}dx$

$I=\frac{1}{4}\int \frac{2\sin x\cos x\cos 2xd\left ( 2x \right )}{\cos x\cos 5x}= \frac{1}{2}\int\frac{\sin 2x\cos 2xd\left ( 2x \right )}{\cos 4x+\cos 6x}= -\frac{1}{2}\int \frac{\cos 2xd\left ( \cos 2x \right )}{2\cos ^{2}x-1+4\cos ^{3}2x-3\cos 2x}$

Đặt $t=\cos 2x$ thì:

$I=-\frac{1}{2}\int \frac{tdt}{4t^{3}+2t^{2}-3t-1}=-\frac{1}{8}\int \left ( -\frac{4}{5} . \frac{dt}{t+1}+\frac{2}{5}\frac{dt}{t-\frac{1+\sqrt{5}}{4}}+ \frac{2}{5}\frac{dt}{t-\frac{1-\sqrt{5}}{4}}\right )$

$=\frac{1}{10}\ln \left | t+1 \right |-\frac{1}{20}\ln \left ( t^{2}+\frac{1}{2} t+\frac{1}{4}\right )+C= \frac{1}{10}\ln \left | \cos 2x+1 \right |-\frac{1}{20}\ln \left ( \cos ^{2} 2x+\frac{1}{2}\cos 2x+\frac{1}{4}\right )+C$