Cho \(\int\limits^{\frac{\pi}{4}}_0\frac{\left(1+\tan x\right)^5}{\cos^2x}dx=\frac{a}{b}\), trong đó a, b là hai số nguyên dương và \(\frac{a}{b}\) là phân số tối giản. Trong các khẳng định sau, khẳng định nào đúng ? \(a< b\) \(ab=1\) \(a-10b=1\) \(a^2+b^2=1\) Hướng dẫn giải: \(\int\limits^{\frac{\pi}{4}}_0\frac{\left(1+\tan x\right)^5}{\cos^2x}dx=\int\limits^{\frac{\pi}{4}}_0\left(1+\tan x\right)^5\text{d}\left(1+\tan x\right)\) (vì \(\left(1+\tan x\right)'=\frac{1}{\cos^2x}\)) \(=\frac{1}{6}\left(1+\tan x\right)^6|^{\frac{\pi}{4}}_0\) \(=\frac{1}{6}\left(2^6-1\right)=\frac{63}{6}=\frac{21}{2}\) => a = 21, b = 2. Do đó A, B, D sai; C đúng. Chọn C.
Trong các khẳng định sau, khẳng định nào sai ? \(\sin\left(\int\limits^{\pi}_0\left(\pi-x\right)\sin xdx\right)=0\) \(\cos\left(\frac{1}{2}\int\limits^{\pi}_0\left(\pi-x\right)\sin xdx\right)=0\) \(tan\left(\frac{3}{4}\int\limits^{\pi}_0\left(\pi-x\right)\sin xdx\right)=-1\) \(\cos\left(2\int\limits^{\pi}_0\left(\pi-x\right)\sin xdx\right)=-1\) Hướng dẫn giải: \(\int\limits^{\pi}_0\left(\pi-x\right)\sin xdx=\int\limits^{\pi}_0\pi\sin x\text{d}x-\int\limits^{\pi}_0x.\sin x\text{d}x\) \(=\pi\left(-\cos x\right)|^{\pi}_0-\int\limits^{\pi}_0x.\sin x\text{d}x\) \(=2\pi-\int\limits^{\pi}_0x.\sin x\text{d}x\) (*) Ta tính \(\int\limits^{\pi}_0x.\sin x\text{d}x\) bằng tích phân từng phần như sau: Đặt \(\begin{cases}u=x\\v'=\sin x\end{cases}\) \(\Rightarrow\begin{cases}u'=1\\v=-\cos x\end{cases}\) => \(\int\limits^{\pi}_0x.\sin x\text{d}x=-x.\cos x|^{\pi}_0+\int\limits^{\pi}_0\cos x\text{d}x\) \(=\pi+\sin x|^{\pi}_0=\pi\) Thay vào (*) ta có: \(\int\limits^{\pi}_0\left(\pi-x\right)\sin xdx=2\pi-\pi=\pi\) Vậy: \(\sin\left(\int\limits^{\pi}_0\left(\pi-x\right)\sin xdx\right)=\sin\pi=0\) \(\cos\left(\frac{1}{2}\int\limits^{\pi}_0\left(\pi-x\right)\sin xdx\right)=\cos\frac{\pi}{2}=0\) \(\tan\left(\frac{3}{4}\int\limits^{\pi}\left(\pi-x\right)\sin xdx_0\right)=\tan\frac{3\pi}{4}=-1\) \(\cos\left(2\int\limits^{\pi}_0\left(\pi-x\right)\sin xdx\right)=\cos2\pi=1\ne-1\). Vậy \(\cos\left(2\int\limits^{\pi}_0\left(\pi-x\right)\sin xdx\right)=-1\) là khẳng định sai.
Tính tích phân \(\sin\left(\int\limits_0^{\pi}\pi x\cos xdx\right)\) \(\sin\left(\int\limits^{\pi}_0\pi x\cos xdx\right)=1\) \(\sin\left(\int\limits^{\pi}_0\pi x\cos xdx\right)=0\) \(\sin\left(\int\limits^{\pi}_0\pi x\cos xdx\right)=\pi\) \(sin\left(\int\limits^{\pi}_0\pi x\cos xdx\right)=\frac{\sqrt{3}}{2}\) Hướng dẫn giải: Ta có: \(\int\limits_0^{\pi}\pi x\cos xdx=\pi\int\limits^{\pi}_0x\cos x\text{d}x\) Đặt \(\begin{cases}u=x\\v'=\cos x\end{cases}\) \(\Rightarrow\begin{cases}u'=1\\v=\sin x\end{cases}\) \(\pi\int\limits_0^{\pi}x\cos xdx=\pi\left(x.\sin x|^{\pi}_0-\int\limits^{\pi}_0\sin x\text{d}x\right)\) \(=\pi\left(x\sin x|^{\pi}_0+\cos x|^{\pi}_0\right)\) \(=-2\pi\) Suy ra: \(\sin\left(\int\limits_0^{\pi}\pi x\cos xdx\right)=\sin\left(-2\pi\right)=0\) . Chọn \(\sin\left(\int\limits^{\pi}_0\pi x\cos xdx\right)=0\).
Trong các khẳng định sau, khẳng định nào sai ? \(\sin\left(\int\limits^1_0\frac{\pi x}{2}e^xdx-\alpha\right)=\cos\alpha,\forall\alpha\) \(\cos\left(\int\limits^1\frac{\pi x}{2}e^xdx-\alpha_0\right)=\sin\alpha,\forall\alpha\) \(\sin\left(\int\limits_0^1\pi xe^xdx-\alpha\right)=\sin\alpha,\forall\alpha\) \(\cos\left(\int\limits_0^1\pi xe^xdx-\alpha\right)=\cos\alpha,\forall\alpha\) Hướng dẫn giải: Ta tính \(I=\int\limits_0^1\pi xe^xdx\) bằng phương pháp tích phân từng phần: Đặt \(\begin{cases}u=x\\v'=e^x\end{cases}\) \(\Rightarrow\begin{cases}u'=1\\v=e^x\end{cases}\) Suy ra: \(I=\pi\left(x.e^x|^1_0-\int\limits^1_0e^x\text{d}x\right)=\pi\left(x.e^x-e^x\right)|^1_0=\pi\) Vậy: \(\sin\left(\int\limits^1_0\frac{\pi x}{2}e^xdx-\alpha\right)=\sin\left(\frac{\pi}{2}-\alpha\right)=\cos\alpha,\forall\alpha\) \(\cos\left(\int\limits^1_0\frac{\pi x}{2}e^xdx-\alpha\right)=\cos\left(\frac{\pi}{2}-\alpha\right)=\sin\alpha,\forall\alpha\) \(\sin\left(\int\limits_0^1\pi xe^xdx-\alpha\right)=\sin\left(\pi-\alpha\right)=\sin\alpha,\forall\alpha\) \(\cos\left(\int\limits_0^1\pi xe^xdx-\alpha\right)=\cos\left(\pi-\alpha\right)=-\cos\alpha,\forall\alpha\) Do đó \(\cos\left(\int\limits_0^1\pi xe^xdx-\alpha\right)=\cos\alpha,\forall\alpha\) là khẳng định sai.
Biết \(\int\limits^1_0\left(\frac{1}{2x+1}-\frac{1}{3x+1}\right)dx=\frac{1}{6}ln\frac{a}{b}\), trong đó a, b là hai số nguyên dương và \(\frac{a}{b}\) là phân số tối giản. Trong các khẳng định sau, khẳng định nào sai ? \(a-b=11\) \(\frac{a}{3}+\frac{b}{4}=7\) \(a+b>22\) \(\sqrt[3]{a}+\sqrt{b}=7\) Hướng dẫn giải: \(\int\limits^1_0\left(\frac{1}{2x+1}-\frac{1}{3x+1}\right)dx=\int\limits^1_0\frac{1}{2x+1}\text{d}x-\int\limits^1_0\frac{1}{3x+1}\text{d}x\) \(=\frac{1}{2}\int\limits^1_0\frac{\text{d}\left(2x+1\right)}{2x+1}-\frac{1}{3}\int\limits^1_0\frac{\text{d}\left(3x+1\right)}{3x+1}\) \(=\left[\frac{1}{2}\ln\left(2x+1\right)-\frac{1}{3}\ln\left(3x+1\right)\right]|^1_0\) \(=\frac{1}{2}\ln3-\frac{1}{3}\ln4\) \(=\frac{3\ln3-2\ln4}{6}=\frac{1}{6}\left(\ln3^3-\ln4^2\right)=\frac{1}{6}\ln\frac{27}{16}\) Vậy a = 27, b = 16 , do đó A, B, D đúng; C sai. Chọn C.
Biết \(F'\left(x\right)=\frac{a\sin^2x\cos^2x+b\sqrt{3}}{\sin^2x\cos^2x};F\left(\frac{\pi}{6}\right)=-\frac{\pi}{2};F\left(\frac{\pi}{4}\right)=\frac{\pi}{4};F\left(\frac{\pi}{3}\right)=\pi\). Tìm hàm số \(F\left(x\right)\) ? \(F\left(x\right)=x+\frac{\pi}{\sqrt{3}}\left(\tan x-\cot x\right)-\frac{\pi}{12}\) \(F\left(x\right)=x+\frac{\pi}{\sqrt{3}}\left(\tan x-\cot x\right)\) \(F\left(x\right)=9x-2\pi\) \(F\left(x\right)=x-\frac{\pi}{\sqrt{3}}\left(\tan x-\cot x\right)+\frac{\pi}{6}\) Hướng dẫn giải: \(F'\left(x\right)=a+b\sqrt{3}\frac{1}{\sin^2x.\cos^2x}=a+b\sqrt{3}\frac{\sin^2x+\cos^2x}{\sin^2x.\cos^2x}\) \(=a+b\sqrt{3}\left(\frac{1}{\cos^2x}+\frac{1}{\sin^2x}\right)\) \(\Rightarrow F\left(x\right)=\int\left[a+b\sqrt{3}\left(\frac{1}{\cos^2x}+\frac{1}{\sin^2x}\right)\right]\text{d}x\) \(=ax+b\sqrt{3}\left(\tan x-\cot x\right)+c\) Ta tìm a, b, c biết: \(\begin{cases}F\left(\frac{\pi}{6}\right)=a.\frac{\pi}{6}+b\sqrt{3}\left(\tan\frac{\pi}{6}-\cot\frac{\pi}{6}\right)+c=-\frac{\pi}{2}\\F\left(\frac{\pi}{4}\right)=a.\frac{\pi}{4}+b\sqrt{3}\left(\tan\frac{\pi}{4}-\cot\frac{\pi}{4}\right)+c=\frac{\pi}{4}\\F\left(\frac{\pi}{3}\right)=a.\frac{\pi}{3}+b\sqrt{3}\left(\tan\frac{\pi}{3}-\cot\frac{\pi}{3}\right)+c=\pi\end{cases}\) \(\Leftrightarrow\begin{cases}a.\frac{\pi}{6}+b\sqrt{3}\left(\frac{1}{\sqrt{3}}-\sqrt{3}\right)+c=-\frac{\pi}{2}\\a.\frac{\pi}{4}+b\sqrt{3}\left(1-1\right)+c=\frac{\pi}{4}\\a.\frac{\pi}{3}+b\sqrt{3}\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)+c=\pi\end{cases}\) Lấy \(a=1\) => \(b=\frac{\pi}{3};c=0\)
Tính \(\int\limits^2_1\frac{lnx}{x^3}dx\) ? \(\int\limits^2_1\frac{lnx}{x^3}dx=\frac{2+ln2}{16}\) \(\int\limits^2_1\frac{lnx}{x^3}dx=\frac{3-2ln2}{16}\) \(\int\limits^2_1\frac{lnx}{x^3}dx=\frac{3+ln2}{16}\) \(\int\limits^2_1\frac{lnx}{x^3}dx=\frac{3+2ln2}{16}\) Hướng dẫn giải: Đặt \(\begin{cases}u=\ln x\\v'=\frac{1}{x^3}\end{cases}\) \(\Rightarrow\begin{cases}u'=\frac{1}{x}\\v=\int x^{-3}\text{d}x=\frac{1}{-3+1}.x^{-3+1}=-\frac{1}{2}.\frac{1}{x^2}\end{cases}\) \(\int\limits^2_1\frac{lnx}{x^3}dx=\ln x.\left(-\frac{1}{2}\frac{1}{x^2}\right)|^2_1-\int\limits^2_1\frac{1}{x}\left(-\frac{1}{2}\frac{1}{x^2}\right)\text{d}x\) \(=-\frac{1}{8}\ln2+\frac{1}{2}\int\limits^2_1x^{-3}\text{d}x\) \(=-\frac{1}{8}\ln2+\frac{1}{2}.\frac{1}{-3+1}x^{-3+1}|^2_1\) \(=-\frac{1}{8}\ln2-\frac{1}{4}.\frac{1}{x^2}|^2_1\) \(=-\frac{1}{8}\ln2-\frac{1}{16}+\frac{1}{4}\) \(=\frac{3-2\ln2}{16}\)
Tính \(\int\limits^1_0\left(x-2\right)e^{2x}dx\) ? \(\int\limits^1_0\left(x-2\right)e^{2x}dx=\frac{5+3e^2}{4}\) \(\int\limits^1_0\left(x-2\right)e^{2x}dx=\frac{-5-3e^2}{4}\) \(\int\limits^1_0\left(x-2\right)e^{2x}dx=\frac{5-3e^2}{4}\) \(\int\limits^1_0\left(x-2\right)e^{2x}dx=\frac{5-3e^2}{2}\) Hướng dẫn giải: Đặt \(\begin{cases}u=x-2\\v'=e^{2x}\end{cases}\) \(\Rightarrow\begin{cases}u'=1\\v=\frac{1}{2}e^{2x}\end{cases}\) \(I=\int\limits^1_0\left(x-2\right)e^{2x}dx=\frac{1}{2}\left(x-2\right)e^{2x}|^1_0-\frac{1}{2}\int\limits^1_0e^{2x}\text{d}x\) \(=\left(\frac{1}{2}\left(x-2\right)e^{2x}-\frac{1}{4}e^{2x}\right)|^1_0\) \(=\left(\frac{1}{2}xe^{2x}-\frac{5}{4}e^{2x}\right)|^1_0\) \(=\frac{1}{2}e^2-\frac{5}{4}e^2+\frac{5}{4}\) \(=\frac{5-3e^2}{4}\)
Tính tích phân \(\int\limits^e_1x^3ln^2xdx\) \(\int\limits^e_1x^3ln^2xdx=\frac{5e^3-1}{32}\) \(\int\limits^e_1x^3ln^2xdx=\frac{5e^2-1}{32}\) \(\int\limits^e_1x^3ln^2xdx=\frac{5e^4-1}{32}\) \(\int\limits^e_1x^3ln^2xdx=\frac{5e-1}{32}\) Hướng dẫn giải: Đặt \(\begin{cases}u=\ln^2x\\v'=x^3\end{cases}\) \(\Rightarrow\begin{cases}u'=\frac{2\ln x}{x}\\v=\frac{x^4}{4}\end{cases}\) => \(I=\int\limits x^3ln^2xdx=\frac{1}{4}x^4\ln^2x-\int\frac{1}{2}x^3\ln x\text{d}x\) Ta lại tính \(I_1=\int\frac{1}{2}x^3\ln x\text{d}x\) bằng phương pháp tích phân từng phần: Đặt \(\begin{cases}u=\ln x\\v'=x^3\end{cases}\) \(\Rightarrow\begin{cases}u'=\frac{1}{x}\\v=\frac{1}{4}x^4\end{cases}\) \(I_1=\frac{1}{2}\left[\frac{1}{4}x^4\ln x-\frac{1}{4}\int x^3\text{d}x\right]\) \(=\frac{1}{2}\left[\frac{1}{4}x^4\ln x-\frac{1}{16}x^4\right]\) Vậy \(I=\frac{1}{4}x^4\ln^2x-\frac{1}{2}\left[\frac{1}{4}x^4\ln x-\frac{1}{16}x^4\right]\) \(\int\limits^e_1x^3ln^2xdx=\left[\frac{1}{4}x^4\ln^2x-\frac{1}{8}x^4\ln x+\frac{1}{32}x^4\right]|^e_1\) \(=\frac{1}{4}e^4\ln^2e-\frac{1}{8}e^4\ln e+\frac{1}{32}e^4-\frac{1}{32}\) \(=\frac{5e^4-1}{32}\)
Tính tích phân \(I=\)\(\int\limits^3_1\frac{3+\ln x}{\left(x+1\right)^2}dx\) . \(I=\dfrac{-3+\ln27-ln16}{4}\) \(I=\dfrac{3+\ln27-ln16}{4}\) \(I=\dfrac{3-\ln27+ln16}{4}\) \(I=\dfrac{3-\ln27-ln16}{4}\) Hướng dẫn giải: Đặt \(\begin{cases}u=3+\ln x\\v'=\frac{1}{\left(x+1\right)^2}\end{cases}\) \(\Rightarrow\begin{cases}u'=\frac{1}{x}\\v=-\frac{1}{x+1}\end{cases}\) \(\int\limits^3_1\frac{3+\ln x}{\left(x+1\right)^2}dx=-\frac{3+\ln x}{x+1}|^3_1+\int\limits^3_1\frac{1}{x\left(x+1\right)}\text{d}x\) \(=-\frac{3+\ln x}{x+1}+\int\limits^3_1\left(\frac{1}{x}-\frac{1}{x+1}\right)\text{d}x\) \(=\left[-\frac{3+\ln x}{x+1}+\ln x-\ln\left(x+1\right)\right]|^3_1\) \(=\left[-\frac{3+\ln x}{x+1}+\ln\frac{x}{x+1}\right]|^3_1\) \(=\frac{3}{4}+\frac{3}{4}\ln3-\ln2\) \(=\frac{3+\ln27-\ln16}{4}\)
