Lời giải:
\(I=\int \frac{1}{2\sin x-\cos x+5}dx\)
Đặt \(t=\tan \frac{x}{2}\)
\(\Rightarrow \sin x=\frac{2t}{t^2+1}; \cos x=\frac{1-t^2}{1+t^2}\)
\(dt=d(\tan \frac{x}{2})=(\tan \frac{x}{2})'dx=\frac{1}{2\cos ^2\frac{x}{2}}dx=\frac{t^2+1}{2}dx\)
\(\Leftrightarrow dx=\frac{2dt}{t^2+1}\)
Do đó: \(I=\int \frac{1}{3t^2+2t+2}dt\)
\(\Leftrightarrow \frac{I}{3}=\int \frac{1}{9t^2+6t+6}dt=\int \frac{1}{(3t+1)^2+5}dt\)
Đặt \(3t+1=\sqrt{5}\tan m\Rightarrow dt=\frac{\sqrt{5}dm}{3\cos^2 m}\)
\(\frac{I}{3}=\int \frac{\sqrt{5}dm}{3}=\frac{m}{3\sqrt{5}}+c\Leftrightarrow I=\frac{m}{\sqrt{5}}+c\)
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