Tích phân \(I=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\left(x^{3}+2 x\right) \cos x+x \cos ^{2} x}{\cos x} d x\) có giá trị là:
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Lời giải:
Báo saiTa có:
\(I=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\left(x^{3}+2 x\right) \cos x+x \cos ^{2} x}{\cos x} d x=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}}\left(x^{3}+2 x\right) d x+\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}} x \cos x d x=\left.\left(\frac{1}{4} x^{4}+x^{2}\right)\right|_{\frac{\pi}{6}} ^{\frac{\pi}{2}}+\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}} x \cos x d x\)
Xét \(I_{1}=\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} x \cos x d x\)
Đặt \(\left\{\begin{array}{l} u=x \\ d v=\cos x d x \end{array} \Rightarrow\left\{\begin{array}{l} d u=d x \\ v=\sin x \end{array}\right.\right.\)
Khi đó:
\(I_{1}=\left.(x \sin x)\right|_{\frac{\pi}{6}} ^{\frac{\pi}{2}}-\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}} \sin x d x=\frac{\pi}{4}-\frac{\sqrt{3}}{2}\)
\(\Rightarrow I=\left.\left(\frac{1}{4} x^{4}+x^{2}\right)\right|_{\frac{\pi}{6}} ^{\frac{\pi}{2}}+I_{1}=\frac{5 \pi^{4}}{324}+\frac{2 \pi^{2}}{9}+\frac{\pi}{4}-\frac{\sqrt{3}}{2}\)