Question

Tính tích phân các hàm lượng giác sau :

a) \(I_1=\int_1^2\left(3x^2+\cos x+\frac{1}{x}\right)dx\)

b) \(I_2=\int_1^2\left(\frac{4}{x}-5x^2+2\sqrt{x}\right)dx\)

c) \(I_3=\int_a^b\frac{\left|x\right|}{x}dx\), với ab>0

d) \(I_5=\int_0^{\frac{\pi}{2a}}\left(x+3\right)\sin ax.dx\) với a>0

e)\(I_4=\int_0^{\pi}\sqrt{\frac{1+\cos2x}{2}}dx\)

Answer 1

\(I_1=3\int_1^2x^2dx+\int_1^2\cos xdx+\int_1^2\frac{dx}{x}=x^3\)\(|^2 _1\)+\(\sin x\)\(|^2_1\) +\(\ln\left|x\right|\)\(|^2_1\)

    \(=\left(8-1\right)+\left(\sin2-\sin1\right)+\left(\ln2-\ln1\right)\)

     \(=7+\sin2-\sin1+\ln2\)

Answer 2

b) \(I_2=4\int_1^2\frac{dx}{x}-5\int_1^2x^4dx+2\int_1^2\sqrt{x}dx\)

         \(=4\left(\ln2-\ln1\right)-\left(2^5-1^5\right)+\frac{4}{3}\left(2\sqrt{2}-1\sqrt{1}\right)\)

         \(=4\ln2+\frac{8\sqrt{2}}{3}-32\frac{1}{3}\)

Answer 3

c) Ta cần xét 2 trường hợp 1) 0<a<b và 2) a<b<0

1) Nếu 0<a<b, khi đó \(f\left(x\right)=\frac{\left|x\right|}{x}=1\) vì \(x>0\) 

Do đó

\(\int_a^bf\left(x\right)dx=\int_a^bdx=b-a\)

2) Nếu a<b<0, khi đó \(f\left(x\right)=\frac{\left|x\right|}{x}=\frac{-x}{x}=1\) vì \(x<0\)

Do đó :

\(\int_a^bf\left(x\right)dx=\int_a^b\left(-1\right)dx=-\left(b-a\right)=a-b\)

 

Answer 4

d) Áp dụng công thức tích phân từng phần \(\int_a^budv=uv^b_a-\int_a^bvdu\) cho \(u=x+3;dv=\sin axdx\)

Ta tìm được \(I_5=\left(-\frac{x+3}{a}\cos ax+\frac{1}{a^2}\sin ax\right)_0^{\frac{\pi}{2a}}\)

                        \(=\frac{1+3a}{a^2}\)

Answer 5

e) \(I_4=\int_0^{\pi}\sqrt{\frac{1+\cos^2x}{2}}dx=\int_0^{\pi}\sqrt{\cos^2x}dx=\int_0^{\pi}\left|\cos x\right|dx\)

Mặt khác ta có \(\left|\cos x\right|=\cos x,0\le x\le\frac{\pi}{2}\) và \(\left|\cos x\right|=-\cos x,\frac{\pi}{2}\le x\le\pi\)

Do đó :

\(I_4=\int_0^{\frac{\pi}{2}}\cos xdx-\int_{^{\frac{\pi}{2}}}^{\pi}\cos xdx\)

    \(=\sin x^{\frac{\pi}{2}}_0-\sin x^{\pi}_{\frac{\pi}{2}}\)

    \(=\left(1-0\right)-\left(0-1\right)=2\)