Tính tích phân sau I= tích phân từ pi/6 đến pi/3 4sin^2 x+!/cosx+ căn 3sinx dx

Trả lời

Giả sử: $4{\sin }^{2}x+1=\left(A\sin x+B\cos x\right)\left(\cos x+\sqrt{3}\sin x\right)+C\left({\sin }^{2}x+{\cos }^{2}x\right)$
$\iff 4{\sin }^{2}x+1=\left(A\sqrt{3}+C\right){\sin }^{2}x+\left(A+B\sqrt{3}\right)\sin x\cos x+\left(B+C\right){\cos }^{2}x$
Đồng nhất hai vế ta có: $\left\{\begin{array}{l}A\sqrt{3}+C=4\\ A+B\sqrt{3}=0\\ B+C=1\end{array}\right.\Rightarrow \left\{\begin{array}{l}A=\sqrt{3}\\ B=-1\\ C=2\end{array}\right.$.
$\begin{array}{l}\Rightarrow I=\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{3}}{\int }}\frac{\left(\sqrt{3}\sin x-\cos x\right)\left(\cos x+\sqrt{3}\sin x\right)+2}{\cos x+\sqrt{3}\sin x}\text{d}x\\ =\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{3}}{\int }}\left(\sqrt{3}\sin x-\cos x\right)\text{d}x+2\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{3}}{\int }}\frac{\text{d}x}{\cos x+\sqrt{3}\sin x}={\left.\left(-\sqrt{3}\cos x-\sin x\right)\right|}_{\frac{\pi }{6}}^{\frac{\pi }{3}}+J=2-\sqrt{3}+J\\ J=2\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{3}}{\int }}\frac{\text{d}x}{\cos x+\sqrt{3}\sin x}=\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{3}}{\int }}\frac{\text{d}x}{\sin \left(x+\frac{\pi }{6}\right)}=\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{3}}{\int }}\frac{\text{d}x}{2\sin \left(\frac{x}{2}+\frac{\pi }{12}\right)\cos \left(\frac{x}{2}+\frac{\pi }{12}\right)}\\ =\frac{1}{2}\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{3}}{\int }}\frac{\text{d}x}{\tan \left(\frac{x}{2}+\frac{\pi }{12}\right){\cos }^2\left(\frac{x}{2}+\frac{\pi }{12}\right)}=\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{3}}{\int }}\frac{d\left[\tan \left(\frac{x}{2}+\frac{\pi }{12}\right)\right]}{\tan \left(\frac{x}{2}+\frac{\pi }{12}\right)}={\left.\ln \left|\tan \left(\frac{x}{2}+\frac{\pi }{12}\right)\right|\right|}_{\frac{\pi }{6}}^{\frac{\pi }{3}}=\frac{1}{2}\ln 3\\ \Rightarrow I=2-\sqrt{3}+\frac{1}{2}\ln 3.\end{array}$