Trong các khẳng định sau, khẳng định nào sai ? \(\left(\frac{1}{\cos x}\right)'=\frac{\sin x}{\cos^2x}\) \(\int\limits^{\frac{\pi}{3}}_0\frac{x\sin x}{\cos^2x}dx=\frac{x}{\cos x}|^{\frac{\pi}{3}}_0-\int\limits^{\frac{\pi}{3}}_0\frac{1}{\cos x}dx\) \(\int\limits^{\frac{\pi}{3}}_0\frac{1}{\cos x}dx=\int\limits^{\frac{\pi}{3}}_0\frac{d\left(\sin x\right)}{\sin^2x}=\frac{1}{2}\ln\left(\frac{1+\sin x}{1-\sin x}\right)|^{\frac{\pi}{3}}_0\) \(\int\limits^{\frac{\pi}{3}}_0\frac{x\sin x}{\cos^2x}dx=\frac{2\pi}{3}-ln\left(2-\sqrt{3}\right)\) Hướng dẫn giải: Ta có: *) \(\left(\frac{1}{\cos x}\right)'=\frac{-\left(\cos x\right)'}{\cos^2x}=\frac{\sin x}{\cos^2x}\), sử dụng công thức này để tính tích phân từng phần trong tính tích phân các câu còn lại. *) Đặt \(\begin{cases}u=x\\v'=\frac{\sin x}{\cos^2x}\end{cases}\) Theo công thức trên suy ra \(\begin{cases}u'=1\\v=\frac{1}{\cos x}\end{cases}\) Vậy \(\int\limits^{\frac{\pi}{3}}_0\frac{x\sin x}{\cos^2x}dx=\frac{x}{\cos x}|^{\frac{\pi}{3}}_0-\int\limits^{\frac{\pi}{3}}_0\frac{1}{\cos x}dx\) Ta lại tính: \(\int\limits^{\frac{\pi}{3}}_0\frac{1}{\cos x}\text{d}x=\int\limits^{\frac{\pi}{3}}_0\frac{\cos x}{\cos^2x}\text{d}x=\int\limits^{\frac{\pi}{3}}_0\frac{\text{d}\left(\sin x\right)}{1-\sin^2x}\) \(=\int\limits^{\frac{\pi}{3}}_0\frac{1}{2}\left[\frac{1}{1-\sin x}+\frac{1}{1+\cos x}\right]\text{d}\left(\sin x\right)\) \(=\) \(\frac{1}{2}\left[-\int\limits^{\frac{\pi}{3}}_0\frac{\text{d}\left(1-\sin x\right)}{1-\sin x}+\int\limits^{\frac{\pi}{3}}_0\frac{\text{d}\left(1+\sin x\right)}{1+\sin x}\right]\) \(=\frac{1}{2}\left[-\ln\left|1-\sin x\right|+\ln\left|1+\sin x\right|\right]|^{\frac{\pi}{3}}_0\) \(=\frac{1}{2}\ln\frac{1+\sin x}{1-\sin x}|^{\frac{\pi}{3}}_0\) Vậy: \(\int\limits^{\frac{\pi}{3}}_0\frac{x\sin x}{\cos^2x}dx=\frac{x}{\cos x}|^{\frac{\pi}{3}}_0-\frac{1}{2}\ln\frac{1+\sin x}{1-\sin x}|^{\frac{\pi}{3}}_0\) \(=\frac{\frac{\pi}{3}}{\frac{1}{2}}-\frac{1}{2}\ln\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\) \(=\frac{2\pi}{3}-\frac{1}{2}\ln3\) Vậy \(\int\limits^{\frac{\pi}{3}}_0\frac{x\sin x}{\cos^2x}dx=\frac{2\pi}{3}-ln\left(2-\sqrt{3}\right)\) là khẳng định sai.
Tính \(I=\int\limits^{ln3}_0\frac{3e^{2x+1}-2}{e^x}\text{d}x\) . \(I=6e-\frac{4}{3}\) \(I=4e+\frac{3}{4}\) \(I=6e+\frac{4}{3}\) \(I=5e-\frac{4}{3}\) Hướng dẫn giải: \(I=\int\limits^{ln3}_0\frac{3e^{2x+1}-2}{e^x}\text{d}x\) \(=\int\limits^{ln3}_0\left(3.e^{x+1}-2.e^{-x}\right)\text{d}x\) \(=3\int\limits^{ln3}_0e^{x+1}\text{d}\left(x+1\right)+2\int\limits^{\ln3}_0e^{-x}\text{d}\left(-x\right)\) \(=\left[3e^{x+1}+2.e^{-x}\right]|^{\ln3}_0\) \(=\left[3e^{\ln3+1}+2.e^{-\ln3}-3e-2\right]\) \(=3e^{\ln3}.e+\frac{2}{e^{\ln3}}-3e-2\) \(=3.3.e+\frac{2}{3}-3e-2\) \(=6e-\frac{4}{3}\) Chọn A.
Tính \(I=\int\limits^{ln2}_0\frac{e^{3x}+1}{e^x+1}dx\) \(I=\frac{1}{2}+ln2\) \(I=\frac{1}{2}-3ln2\) \(I=\frac{1}{2}+2ln2\) \(I=-\frac{1}{2}-ln2\) Hướng dẫn giải: \(I=\int\limits^{ln2}_0\frac{e^{3x}+1}{e^x+1}dx\) \(=\int\limits^{\ln2}_0\frac{\left(e^x\right)^3+1}{e^x+1}\text{d}x\) \(=\int\limits^{\ln2}_0\frac{\left(e^x+1\right)\left(x^{2x}-e^x+1\right)}{e^x+1}\text{d}x\) \(=\int\limits^{\ln2}_0\left(e^{2x}-e^x+1\right)\text{d}x\) \(=\frac{1}{2}\int\limits^{\ln2}_0e^{2x}\text{d}\left(2x\right)-\int\limits^{\ln2}_0e^x\text{d}x+\int\limits^{\ln2}_0\text{d}x\) \(=\left(\frac{1}{2}e^{2x}-e^x+x\right)|^{\ln2}_0\) \(=\frac{1}{2}e^{2\ln2}-e^{\ln2}+\ln2-\frac{1}{2}+1-0\) \(=\frac{1}{2}e^{\ln2^2}-2+\ln2-\frac{1}{2}+1\) \(=\frac{1}{2}.2^2-2+\ln2-\frac{1}{2}+1\) \(=\frac{1}{2}+\ln2\) Chọn A.
Tính \(I=\int\limits^e_0\frac{1}{\sqrt{x+1}+\sqrt{x}}dx\) \(I=\frac{1}{\sqrt{e+1}+\sqrt{x}}-2\) \(I=2\left(\frac{1}{\sqrt{e+1}+\sqrt{x}}-1\right)\) \(I=\frac{2}{3}\left(\left(e+1\right)\sqrt{e+1}-e\sqrt{e}-1\right)\) \(I=\frac{2}{3}\left(\left(e+1\right)\sqrt{e+1}-e\sqrt{e}+1\right)\) Hướng dẫn giải: \(I=\int\limits^e_0\frac{1}{\sqrt{x+1}+\sqrt{x}}dx\) \(=\int\limits^e_0\frac{\sqrt{x+1}-\sqrt{x}}{\left(\sqrt{x+1}+\sqrt{x}\right)\left(\sqrt{x+1}-\sqrt{x}\right)}\text{d}x\) \(=\int\limits^e_0\left(\sqrt{x+1}-\sqrt{x}\right)\text{d}x\) \(=\int\limits^e_0\left[\left(x+1\right)^{\frac{1}{2}}-x^{\frac{1}{2}}\right]\text{d}x\) \(=\int\limits^e_0\left(x+1\right)^{\frac{1}{2}}\text{d}\left(x+1\right)-\int\limits^e_0x^{\frac{1}{2}}\text{d}x\) \(=\left[\frac{1}{1+\frac{1}{2}}\left(x+1\right)^{1+\frac{1}{2}}-\frac{1}{1+\frac{1}{2}}x^{1+\frac{1}{2}}\right]|^e_0\) \(=\left[\frac{2}{3}\left(x+1\right)^{\frac{3}{2}}-\frac{2}{3}x^{\frac{3}{2}}\right]|^e_0\) \(=\frac{2}{3}\left[\left(e+1\right)\sqrt{e+1}-e\sqrt{e}-1\right]\) Chọn C.
Giải phương trình ẩn a sau đây : \(\int\limits^a_0\cos xdx=0\) \(a=\frac{\pi}{3}\) \(a=\frac{\pi}{3}+k2\pi,k\in\mathbb{Z}\) \(a=\frac{\pi}{6}+k2\pi,k\in\mathbb{Z}\) \(a=k\pi,k\in\mathbb{Z}\) Hướng dẫn giải: \(\int\limits^a_0\cos xdx=0\) \(\Leftrightarrow\sin x|^a_0=0\) \(\Leftrightarrow\sin a-\sin0=0\) \(\Leftrightarrow\sin a=0\) \(\Leftrightarrow a=k\pi,k\in\mathbb{Z}\) Chọn D.
Biết \(\int\limits^{\frac{\pi}{2}}_0\left(e^{\sin x}+\cos x\right)\cos xdx=e-1+a\). Trong các khẳng định sau, khẳng định nào sai ? \(\sin\left(\frac{3\pi}{4}+a-\alpha\right)=-\sin\alpha,\forall\alpha\) \(\cos\left(\frac{3\pi}{4}+a-\alpha\right)=-\cos\alpha,\forall\alpha\) \(\tan\left(\frac{3\pi}{4}+a-\alpha\right)=-\tan\alpha,\forall\alpha\) \(\cot\left(\frac{3\pi}{4}+a-\alpha\right)=-\cot\alpha,\forall\alpha\) Hướng dẫn giải: \(\int\limits^{\frac{\pi}{2}}_0\left(e^{\sin x}+\cos x\right)\cos xdx=\int\limits^{\frac{\pi}{2}}_0e^{\sin x}\cos x\text{d}x+\int\limits^{\frac{\pi}{2}}_0\cos^2x\text{d}x\) \(=\int\limits^{\frac{\pi}{2}}_0e^{\sin x}d\left(\sin x\right)+\int\limits^{\frac{\pi}{2}}_0\frac{1+\cos2x}{2}\text{d}x\) \(=e^{\sin x}|^{\frac{\pi}{2}}_0+\frac{1}{2}x|^{\frac{\pi}{2}}_0+\frac{1}{4}\int\limits^{\frac{\pi}{2}}_0\cos2x\text{d}\left(2x\right)\) \(=\left(e^{\sin x}+\frac{1}{2}x+\frac{1}{4}\sin2x\right)|^{\frac{\pi}{2}}_0\) \(=e+\frac{\pi}{4}-1\) Suy ra \(e+\frac{\pi}{4}-1=e-1+a\) \(\Rightarrow a=\frac{\pi}{4}\) Vậy: \(\frac{3\pi}{4}+a=\frac{3\pi}{4}+\frac{\pi}{4}=\pi\) Góc \(\frac{3\pi}{4}+a-\alpha\) và góc \(\alpha\) là hai góc bù nhau. Suy ra phát biểu \(\sin\left(\frac{3\pi}{4}+a-\alpha\right)=-\sin\alpha,\forall\alpha\) là sai. Chọn A.
Tính \(\int\limits^{\frac{\pi}{4}}_0\frac{a-2a\sin^2x}{1+\sin2x}\text{d}x\) trong đó \(a\) là một số đã cho : \(\int\limits^{\frac{\pi}{4}}_0\frac{a-2a\sin^2x}{1+\sin2x}dx=2a-a\sqrt{2}\) \(\int\limits^{\frac{\pi}{4}}_0\frac{a-2a\sin^2x}{1+\sin2x}dx=\frac{a\sqrt{2}}{2}-1\) \(\int\limits^{\frac{\pi}{4}}_0\frac{a-2a\sin^2x}{1+\sin2x}dx=ln\sqrt{2^a}\) \(\int\limits^{\frac{\pi}{4}}_0\frac{a-2a\sin^2x}{1+\sin2x}dx=\frac{1}{2}lna\) Hướng dẫn giải: \(\int\limits^{\frac{\pi}{4}}_0\frac{a-2a\sin^2x}{1+\sin2x}\text{d}x=a\int\limits^{\frac{\pi}{4}}_0\frac{1-2\sin^2x}{1+\sin2x}\text{d}x\) \(=a\int\limits^{\frac{\pi}{4}}_0\frac{\cos2x}{1+\sin2x}\text{d}x\) \(=\frac{a}{2}\int\limits^{\frac{\pi}{4}}_0\frac{\text{d}\left(1+\sin2x\right)}{1+\sin2x}\) \(=\frac{a}{2}\ln\left|1+\sin2x\right||^{\frac{\pi}{4}}_0\) \(=\frac{a}{2}\ln2\) \(=\ln\sqrt{2^a}\) Chọn C.
Tính tích phân \(\int\limits^{\frac{\pi}{2}}_0\left(1-\cos x\right)^n\sin xdx\) : \(\int\limits^{\frac{\pi}{2}}_0\left(1-\cos x\right)^n\sin xdx=\frac{1}{2n}\) \(\int\limits^{\frac{\pi}{2}}_0\left(1-\cos x\right)^n\sin xdx=\frac{1}{n+1}\) \(\int\limits^{\frac{\pi}{2}}_0\left(1-\cos x\right)^n\sin xdx=\frac{1}{n-1}\) \(\int\limits^{\frac{\pi}{2}}_0\left(1-\cos x\right)^n\sin xdx=\frac{1}{2n-1}\) Hướng dẫn giải: \(\int\limits^{\frac{\pi}{2}}_0\left(1-\cos x\right)^n\sin xdx=\int\limits^{\frac{\pi}{2}}_0\left(1-\cos x\right)^n\text{d}\left(1-\cos x\right)\) \(=\frac{\left(1-\cos x\right)^{n+1}}{n+1}|^{\frac{\pi}{2}}_0\) \(=\frac{1}{n+1}\) Chọn B.
Trong các giá trị sau của n cho sau đây, tìm n để \(\int\limits^{\dfrac{\pi}{3}}_0\cos^nx\sin xdx=\dfrac{15}{64}\) \(n=1\) \(n=2\) \(n=3\) \(n=4\) Hướng dẫn giải: \(\int\limits^{\frac{\pi}{3}}_0\cos^nx\sin xdx=-\int\limits^{\frac{\pi}{3}}_0\cos^nx\text{d}\left(\cos x\right)\) \(=-\frac{\cos^{n+1}x}{n+1}|^{\frac{\pi}{3}}_0=-\left(\frac{\left(\frac{1}{2}\right)^{n+1}}{n+1}-\frac{1}{n+1}\right)\) \(=\frac{1}{n+1}-\frac{1}{2^{n+1}\left(n+1\right)}=\frac{2^{n+1}-1}{2^{n+1}\left(n+1\right)}\) Theo yêu cầu ta có: \(\frac{2^{n+1}-1}{2^{n+1}\left(n+1\right)}=\frac{15}{64}\) => n = 3. Chọn C.
Biết \(\int\limits^1_0\frac{\left(3x-1\right)dx}{x^2+6x+9}=3ln\frac{a}{b}-\frac{5}{6}\), trong đó a, b nguyên dương và \(\frac{a}{b}\) là phân số tối giản Hãy tính tích ab . \(ab=-5\) \(ab=12\) \(ab=6\) \(ab=\frac{5}{4}\) Hướng dẫn giải: \(\frac{3x-1}{x^2+6x+9}=3.\frac{x-\frac{1}{3}}{\left(x+3\right)^2}=3.\left[\frac{x+3-\frac{10}{3}}{\left(x+3\right)^2}\right]\) \(=\frac{3}{x+3}-\frac{10}{\left(x+3\right)^2}\) Vậy \(\int\limits^1_0\frac{\left(3x-1\right)dx}{x^2+6x+9}=\int\limits^1_0\frac{3}{x+3}\text{d}x-\int\limits^1_0\frac{10}{\left(x+3\right)^2}\text{d}x\) \(=3.\ln\left(x+3\right)|^1_0-10.\frac{1}{-2+1}.\left(x+3\right)^{-2+1}|^1_0\) \(=\left[3\ln\left(x+3\right)+\frac{10}{\left(x+3\right)}\right]|^1_0\) \(=3\left(\ln4-\ln3\right)+10\left(\frac{1}{4}-\frac{1}{3}\right)\) \(=3\ln\frac{4}{3}-\frac{5}{6}\) Suy ra \(a=4;b=3\) . Từ đó \(ab=12\). Vậy chọn B.
