Giải bài tập 3 trang 20

Đề bài

Tính các tích phân sau:

a) $\underset{-2}{\overset{4}{\int }}\left(x+1\right)\left(x-1\right)dx$

b) $\underset{1}{\overset{2}{\int }}\frac{x^2-2x+1}{x}dx$

c) $\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(3\sin x-2\right)dx$

d) $\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^{2}x}{1+\cos x}dx$

Lời giải chi tiết

a) $\underset{-2}{\overset{4}{\int }}\left(x+1\right)\left(x-1\right)dx=\underset{-2}{\overset{4}{\int }}\left(x^2-1\right)dx={\left(\frac{x^3}{3}-x\right)|}_{-2}^4=\left(\frac{4^3}{3}-4\right)-\left(\frac{{\left(-2\right)}^3}{3}-\left(-2\right)\right)=18$

b) $\underset{1}{\overset{2}{\int }}\frac{x^2-2x+1}{x}dx=\underset{1}{\overset{2}{\int }}\left(x-2+\frac{1}{x}\right)dx={\left(\frac{x^2}{2}-2x+\ln \left|x\right|\right)|}_1^2$

$=\left(\frac{2{}^2}{2}-2.2+\ln \left|2\right|\right)-\left(\frac{1{}^2}{2}-1.2+\ln \left|1\right|\right)=\ln 2-\frac{1}{2}$

c) $\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(3\sin x-2\right)dx=3\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sin xdx-2\underset{0}{\overset{\frac{\pi }{2}}{\int }}dx=3{\left(-\cos x\right)|}_0^{\frac{\pi }{2}}-2{\left(x\right)|}_0^{\frac{\pi }{2}}$

$=3\left[\left(-\cos \frac{\pi }{2}\right)-\left(-\cos 0\right)\right]-2\left(\frac{\pi }{2}-0\right)=3-\pi$

d) $\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^{2}x}{1+\cos x}dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1-{\cos }^{2}x}{1+\cos x}dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\left(1-\cos x\right)\left(1+\cos x\right)}{1+\cos x}dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1-\cos x\right)dx$

$={\left(x-\sin x\right)|}_0^{\frac{\pi }{2}}=\left(\frac{\pi }{2}-\sin \frac{\pi }{2}\right)-\left(0-\sin 0\right)=\frac{\pi }{2}-1$