Tích phân \( \int\limits_{0}^{\pi}(3 x+2) \cos ^{2} x d x\) bằng:
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Lời giải:
Báo saiĐặt \(I=\int\limits_0^{\pi}(3 x+2) \cos ^{2} x d x\) ta có:
\(I=\frac{1}{2} \int_{0}^{\pi}(3 x+2)(1+\cos 2 x) \mathrm{d} x=\frac{1}{2}\left[\int_{0}^{\pi}(3 x+2) \mathrm{d} x+\int_{0}^{\pi}(3 x+2) \cos 2 x \mathrm{d} x\right]=\frac{1}{2}\left(I_{1}+I_{2}\right)\)
\(I_{1}=\int_{0}^{\pi}(3 x+2) \mathrm{d} x=\left.\left(\frac{3}{2} x^{2}+2 x\right)\right|_{0} ^{\pi}=\frac{3}{2} \pi^{2}+2 \pi\)
\(I_{2}=\int_{0}^{\pi}(3 x+2) \cos 2 x \mathrm{d} x\).
Dùng tích phân từng phần
\(\left\{\begin{array}{l} u=3 x+2 \\ \mathrm{d} v=\cos 2 x \mathrm{d} x \end{array} \Rightarrow\left\{\begin{array}{l} \mathrm{d} u=3 \mathrm{d} x \\ v=\frac{1}{2} \sin 2 x \end{array}\right.\right.\)
Khi đó: \(I_{2}=\left.\frac{1}{2}(3 x+2) \sin 2 x\right|_{0} ^{\pi}-\frac{3}{2} \int_{0}^{\pi} \sin 2 x \mathrm{d} x=0+\left.\frac{3}{4}(\cos 2 x)\right|_{0} ^{\pi}=0\)
Vậy \( I=\frac{1}{2}\left(\frac{3}{2} \pi^{2}+2 \pi\right)=\frac{3}{4} \pi^{2}+\pi\)
