Xét tích phân \(I=\int\limits^2_1\frac{xdx}{1+\sqrt{x-1}}\) và đặt \(t=\sqrt{x-1}\). Trong các khẳng định sau, khẳng định nào sai ? \(dx=2tdt\) \(I=\int\limits^1_0\frac{2t^3+2t}{t+1}dt\) \(I=\int\limits^1_0\left(2t^2-2t+4-\frac{4}{t+1}\right)dt\) \(I=\frac{7}{3}-3ln2\) Hướng dẫn giải: Ta có: \(t=\sqrt{x-1}\Rightarrow x=t^2+1;\text{d}x=2t\text{d}t\) Đổi cận: \(x|^2_1\Rightarrow t|^1_0\) Suy ra: \(I=\int\limits^1_0\frac{t^2+1}{1+t}2t\text{d}t\) \(=\int\limits^1_0\frac{2t^3+2t}{t+1}\text{d}t\) \(=\int\limits^1_0\frac{2t^3+2t^2-2t^2-2t+4t+4-4}{t+1}\text{d}t\) \(=\int\limits^1_0\frac{2t^2\left(t+1\right)-2t\left(t+1\right)+4\left(t+1\right)-4}{t+1}\text{d}t\) \(=\int\limits^1_0\left(2t^2-2t+4-\frac{4}{t+1}\right)\text{d}t\) \(=\left[2\frac{1}{3}t^3-2.\frac{1}{2}t^2+4t-4\ln\left(t+1\right)\right]|^1_0\) \(=\left[\frac{2}{3}t^3-t^2+4t-4\ln\left(t+1\right)\right]|^1_0\) \(=\left[\frac{2}{3}.1^3-1^2+4.1-4\ln\left(1+1\right)\right]\) \(=\frac{11}{3}-4\ln2\) Vậy \(I=\frac{7}{3}-3ln2\) là sai.
Đặt \(I=\int\limits^6_{3\sqrt{2}}\frac{dx}{x\sqrt{x^2-9}}\) và \(x=\frac{3}{\cos t}\) . Trong các khẳng sau, khẳng định nào sai ? \(dx=\frac{3sint}{\cos^2t}dt\) \(\frac{dx}{x\sqrt{x^2-9}}=\frac{sintdt}{3\cos ttant}\) \(I=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{sintdt}{3\cos ttant}\) \(I=\frac{\pi}{36}\) Hướng dẫn giải: Đặt \(x=\frac{3}{\cos t}\) suy ra \(\text{d}x=\dfrac{-3.\left(-\sin t\right)}{\cos^2t}\text{d}t=\dfrac{3\sin t}{\cos^2t}\text{d}t\) Đổi cận: \(x|^6_{3\sqrt{2}}\Rightarrow t|^{\frac{\pi}{3}}_{\frac{\pi}{4}}\) (vì \(\frac{3}{\cos t}=3\sqrt{2}\Rightarrow t=\frac{\pi}{4};\frac{3}{\cos t}=6\Rightarrow t=\frac{\pi}{3}\)) Vậy: \(I=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{\frac{3\sin t}{\cos^2t}\text{d}t}{\frac{3}{\cos t}\sqrt{\frac{9}{\cos^2t}-9}}\) \(=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{\sin t}{3\cos t.\tan t}\text{d}t\) \(=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{1}{3}\text{d}t\) \(=\frac{1}{3}t|^{\frac{\pi}{3}}_{\frac{\pi}{4}}\) \(=\frac{1}{3}\left(\frac{\pi}{3}-\frac{\pi}{4}\right)=\frac{\pi}{36}\) Chú ý: \(\frac{dx}{x\sqrt{x^2-9}}=\frac{sintdt}{3\cos t.\left|tant\right|}\) và \(\frac{dx}{x\sqrt{x^2-9}}=\frac{sintdt}{3\cos ttant}\) trong ngữ cảnh \(x\in\left[\frac{\pi}{4};\frac{\pi}{3}\right]\)
Đặt \(I=\int\limits^2_0\frac{dx}{4+x^2}\) và \(x=2\tan t\) Trong các khẳng định sau, khẳng định nào sai ? \(4+x^2=4\left(1+\tan^2t\right)\) \(dx=2\left(1+\tan^2t\right)dt\) \(I=\int\limits^{\frac{\pi}{4}}_0\frac{1}{2}dt\) \(I=\frac{3\pi}{4}\) Hướng dẫn giải: Đặt \(x=2\tan t\) suy ra \(\text{d}x=2.\frac{1}{\cos^2t}\text{d}t=2\left(1+\tan^2t\right)\text{d}t\) Đổi cận: \(x|^2_0\Rightarrow t|^{\frac{\pi}{4}}_0\) Thay vào ta có: \(I=\int\limits^{\frac{\pi}{4}}_0\frac{2\left(1+\tan^2t\right)\text{d}t}{4+4\tan^2t}\) \(=\int\limits^{\frac{\pi}{4}}_0\frac{1}{2}\text{d}t\) \(=\frac{1}{2}t|^{\frac{\pi}{4}}_0=\frac{1}{2}.\frac{\pi}{4}=\frac{\pi}{8}\)
Xét tích phân \(I=\int\limits^8_3\frac{xdx}{1+\sqrt{x+1}}\). Nếu đặt \(t=\sqrt{x+1}\) thì khẳng định nào trong các khẳng định sau đúng ? \(I=\int\limits^3_2\left(1-t^2\right)dt\) \(I=2\int\limits^3_2\left(t^2-t\right)dt\) \(I=2\int\limits^8_3\left(t-t^2\right)dt\) \(I=\int\limits^3_2\left(t+t^2\right)dt\) Hướng dẫn giải: Đặt \(t=\sqrt{x+1}\) suy ra \(x=t^2-1\) \(\Rightarrow\text{d}x=2t\text{d}t\) Đổi cận \(x|^8_3\Rightarrow t|^3_2\) Vậy \(I=\int\limits^3_2\frac{\left(t^2-1\right).2t\text{d}t}{1+t}\) \(=\int\limits^3_2\frac{\left(t-1\right)\left(t+1\right)2t}{t+1}\text{d}t\) \(=2\int\limits^3_2\left(t^2-t\right)\text{d}t\)
Trong các khẳng định sau, khẳng định nào sai ? Với \(t=\sqrt{4-3\cos x}\) thì \(\cos x=\frac{4-t^2}{3}\) và \(\sin xdx=\frac{2tdt}{3}\) Nếu đặt \(t=\sqrt{4-3\cos x}\) thì \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{\cos x+\sqrt{4-3\cos x}}dx=\frac{2}{5}\int\limits^2_1\left(\frac{4}{4-t}-\frac{1}{1+t}\right)dt\) \(\int\left(\frac{4}{4-t}-\frac{1}{1+t}\right)dt=-\frac{2}{5}\left(4ln\left(t-4\right)+ln\left(t+1\right)\right)\) \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{\cos x+\sqrt{4-3\cos x}}dx=\frac{6}{5}ln\frac{3}{2}\) Hướng dẫn giải: *) \(t=\sqrt{4-3\cos x}\) \(\Leftrightarrow t^2=4-3\cos x\) \(\Leftrightarrow\cos x=\frac{4-t^2}{3}\) Lấy vi phân hai vế: \(-\sin x\text{d}x=-\frac{2}{3}t\text{.d}t\) \(\sin x\text{d}x=\frac{2}{3}t\text{.d}t\) *) Để đổi biến số trong tích phân, ta đổi cận: \(x|^{\frac{\pi}{2}}_0\Rightarrow t|^2_1\) \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{\cos x+\sqrt{4-3\cos x}}\text{d}x=\int\limits^2_1\frac{\frac{2}{3}t.\text{d}t}{\frac{4-t^3}{3}+t}\) \(=\int\limits^2_1\frac{2t}{-t^2+3t+4}\text{d}t\) \(=\int\limits^2_1\frac{2t}{-\left(t+1\right)\left(t-4\right)}\text{d}t\) \(=2\int\limits^2_1\frac{t}{\left(t+1\right)\left(t-4\right)}\text{d}t\) \(=2\int\limits^2_1\frac{t}{5}\left[\frac{1}{4-t}+\frac{1}{t+1}\right]\text{d}t\) \(=\frac{2}{5}\int\limits^2_1t\left[\frac{1}{4-t}+\frac{1}{t+1}\right]\text{d}t\) \(=\frac{2}{5}\int\limits^2_1\left(\frac{t}{4-t}+\frac{t}{t+1}\right)\text{d}t\) \(=\frac{2}{5}\int\limits^2_1\left(\frac{-4+t+4}{4-t}+\frac{t+1-1}{t+1}\right)\text{d}t\) \(=\frac{2}{5}\int\limits^2_1\left(-1+\frac{4}{t-4}+1-\frac{1}{t+1}\right)\text{d}t\) \(=\frac{2}{5}\int\limits^2_1\left(\frac{4}{4-t}-\frac{1}{t+1}\right)\text{d}t\) \(=\frac{2}{5}\left(-4.\ln\left|4-t\right|-\ln\left|t+1\right|\right)|^2_1\) \(=-\frac{2}{5}\left[4.\ln\left(4-t\right)+\ln\left(t+1\right)\right]|^2_1\) \(=-\frac{2}{5}\left[4\ln2+\ln3-4\ln3-\ln2\right]\) \(=-\frac{2}{5}\left[3\ln2-3\ln3\right]\) \(=-\frac{2}{5}.3.\ln\frac{2}{3}\) \(=\frac{6}{5}\ln\frac{3}{2}\) Vậy tất cả công thức trên đều đúng, trừ công thức sau: \(\int\left(\frac{4}{4-t}-\frac{1}{1+t}\right)dt=-\frac{2}{5}\left(4ln\left(t-4\right)+ln\left(t+1\right)\right)\) Vì \(\int\left(\frac{4}{4-t}-\frac{1}{1+t}\right)dt=-\frac{2}{5}\left(4ln\left|4-t\right|+ln\left|t+1\right|\right)\) Trong bài toán tích phân xác định ở trên, vì \(x\in\left[1;2\right]\) nên mới bỏ dấu giá trị tuyệt đối được.
Biết \(\int\limits^3_1\frac{dx}{e^x-1}=ln\left(ae^2+e+b\right)-2\). Khẳng định nào đúng ? \(a+b=2\) \(a+b< 2\) \(a+b>2\) \(ab=2\) Hướng dẫn giải: Đặt \(t=e^x-1\) => \(\text{d}t=e^x\text{d}x\Rightarrow\text{d}x=\frac{\text{d}t}{e^x}=\frac{\text{d}t}{t+1}\) Đổi cận: \(x|^3_1\Rightarrow t|^{e^3-1}_{e-1}\) \(\int\limits^3_1\frac{1}{e^x-1}\text{d}x=\int\limits^{e^3-1}_{e-1}\frac{\text{d}t}{t\left(t+1\right)}\) \(=\int\limits^{e^3-1}_{e-1}\left[\frac{1}{t}-\frac{1}{t+1}\right]\text{d}t\) \(=\left[\ln\left|t\right|-\ln\left|t+1\right|\right]|^{e^3-1}_{e-1}\) \(=\ln\frac{t}{t+1}|^{e^3-1}_{e-1}\) \(=\ln\frac{e^3-1}{e^3}-\ln\frac{e-1}{e}\) \(=\ln\left(\frac{e^3-1}{e^3}:\frac{e-1}{e}\right)\) \(=\ln\frac{e^2+e+1}{e^2}\) \(=\ln\left(e^2+e+1\right)-2\) Vậy ta có: \(\ln\left(e^2+e+1\right)-2=\ln\left(ae^2+e+b\right)-2\) \(\Leftrightarrow e^2\left(1-a\right)=b-1\) \(\Leftrightarrow\begin{cases}a=1\\b=1\end{cases}\) => a + b = 2
Tính tích phân \(\int\limits^{\frac{\pi}{4}}_0\frac{\sin x-\cos x}{\left(1+\sin x+\cos x\right)^2}dx\) \(\int\limits^{\frac{\pi}{4}}_0\frac{\sin x-\cos x}{\left(1+\sin x+\cos x\right)^2}dx=-\frac{3}{2}+\sqrt{2}\) \(\int\limits^{\frac{\pi}{4}}_0\frac{\sin x-\cos x}{\left(1+\sin x+\cos x\right)^2}dx=-1+\sqrt{2}\) \(\int\limits_0^{\frac{\pi}{4}}\frac{\sin x-\cos x}{\left(1+\sin x+\cos x\right)^2}dx=1+\sqrt{2}\) \(\int\limits^{\frac{\pi}{4}}_0\frac{\sin x-\cos x}{\left(1+\sin x+\cos x\right)^2}dx=\sqrt{2}\) Hướng dẫn giải: Đặt \(t=1+\sin x+\cos x\) \(\Rightarrow\text{d}t=\left(\cos x-\sin x\right)\text{d}x\) Đổi cận: \(x|^{\frac{\pi}{4}}_0\Rightarrow t|^{1+\sqrt{2}}_2\) Ta có: \(\int\limits^{\frac{\pi}{4}}_0\frac{\sin x-\cos x}{\left(1+\sin x+\cos x\right)^2}dx=\int\limits^{1+\sqrt{2}}_2\frac{-\text{d}t}{t^2}\) \(=\frac{1}{t}|^{1+\sqrt{2}}_2=\frac{1}{1+\sqrt{2}}-\frac{1}{2}\) \(=\frac{1-\sqrt{2}}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}-\frac{1}{2}=-\frac{3}{2}+\sqrt{2}\)
Tính tích phân \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin2x\cos xdx}{1+\cos x}\) \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin2x\cos xdx}{1+\cos x}=-1+ln2\) \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin2x\cos xdx}{1+\cos x}=-1+3ln2\) \(\int^{\frac{\pi}{2}}_0\frac{\sin2x\cos xdx}{1+\cos x}=-1+2ln2\) \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin2x\cos xdx}{1+\cos x}=2+2ln2\) Hướng dẫn giải: \(I=\) \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin2x\cos xdx}{1+\cos x}=\int\limits^{\frac{\pi}{2}}_0\frac{2\sin x\cos x.\cos x}{1+\cos x}\text{d}x\) \(=\int\limits^{\frac{\pi}{2}}_0\frac{2\cos^2x}{1+\cos x}\sin x\text{d}x\) Đặt \(t=1+\cos x\) \(\Rightarrow\text{d}t=-\sin x\text{d}x\) ; \(\cos^2x=\left(1-t\right)^2\) Đổi cận \(x|^{\frac{\pi}{2}}_0\Rightarrow t|^1_2\), tích phân đã cho bằng: \(I=-\int\limits^1_2\frac{2\left(1-t\right)^2}{t}\text{d}t\) \(=2\int\limits^2_1\frac{1-2t+t^2}{t}\text{d}t\) \(=2\int\limits^2_1\left(\frac{1}{t}-2+t\right)\text{d}t\) \(=2\left(\ln t-2t+\frac{t^2}{2}\right)|^2_1\) \(=2\left(\ln2-4+2+2-\frac{1}{2}\right)\) \(=2\ln2-1\)
Tính tích phân \(\int\limits^{\frac{\pi}{6}}_0\frac{dx}{\cos2x}\) \(\int\limits^{\frac{\pi}{6}}_0\frac{dx}{\cos2x}=\frac{1}{2}ln\left(2-\sqrt{3}\right)\) \(\int\limits^{\frac{\pi}{6}}_0\frac{dx}{\cos2x}=ln\left(2-\sqrt{3}\right)\) \(\int\limits^{\frac{\pi}{6}}_0\frac{dx}{\cos2x}=ln\sqrt{2+\sqrt{3}}\) \(\int\limits^{\frac{\pi}{6}}_0\frac{dx}{\cos2x}=\frac{1}{3}ln\left(2+\sqrt{3}\right)\) Hướng dẫn giải: \(I=\int\limits^{\frac{\pi}{6}}_0\frac{dx}{\cos2x}=\frac{1}{2}\int\limits^{\frac{\pi}{6}}_0\frac{2\cos2x\text{d}x}{\cos^22x}=\frac{1}{2}\int\limits^{\frac{\pi}{6}}_0\frac{2\cos2x\text{d}x}{1-\sin^22x}\) Đặt \(t=\sin2x\) => \(\text{d}t=2.\cos2x\text{d}x\) Đổi cận: \(x|^{\frac{\pi}{6}}_0\Rightarrow t|^{\frac{\sqrt{3}}{2}}_0\) \(I=\frac{1}{2}\int\limits^{\frac{\sqrt{3}}{2}}_0\frac{\text{d}t}{1-t^2}=\frac{1}{2}\int\limits^{\frac{\sqrt{3}}{2}}_0\frac{1}{\left(1-t\right)\left(1+t\right)}\text{d}t\) \(=\frac{1}{4}\int\limits^{\frac{\sqrt{3}}{2}}_0\left(\frac{1}{1-t}+\frac{1}{1+t}\right)\text{d}t\) \(=\frac{1}{4}\left[-\int\limits^{\frac{\sqrt{3}}{2}}_0\frac{\text{d}\left(1-t\right)}{1-t}+\int\limits^{\frac{\sqrt{3}}{2}}_0\frac{\text{d}\left(1+t\right)}{1+t}\right]\) \(=\frac{1}{4}\left(-\ln\left|1-t\right|+\ln\left|1+t\right|\right)|^{\frac{\sqrt{3}}{2}}_0\) \(=\frac{1}{4}\ln\left|\frac{1+t}{1-t}\right||^{\frac{\sqrt{3}}{2}}_0\) \(=\frac{1}{2}\ln\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{1}{2}\ln\left(2+\sqrt{3}\right)^2=\frac{1}{2}\ln\left(2+\sqrt{3}\right)=\ln\sqrt{2+\sqrt{3}}\)
Tính tích phân \(\int\limits^4_0\frac{dx}{\sqrt{2x+1}+1}\) \(\int\limits^2_0\frac{dx}{\sqrt{2x+1}+1}=2-ln3\) \(\int\limits^2_0\frac{dx}{\sqrt{2x+1}+1}=2-2ln2\) \(\int\limits^2_0\frac{dx}{\sqrt{2x+1}+1}=2-ln2\) \(\int\limits^2_0\frac{dx}{\sqrt{2x+1}+1}=4-ln2\) Hướng dẫn giải: Đặt \(t=\sqrt{2x+1}+1\) \(\Rightarrow x=\frac{\left(t-1\right)^2-1}{2}\) \(\text{d}x=\left(t-1\right)\text{d}t\) Đổi cận: \(x|^4_0\Rightarrow t|^4_2\) \(I=\int\limits^4_0\frac{dx}{\sqrt{2x+1}+1}=\int\limits^4_2\frac{t-1}{t}\text{d}t\) \(=\int\limits^4_2\left(1-\frac{1}{t}\text{d}t\right)\) \(=\left(t-\ln t\right)|^4_2=\left(4-\ln4-2+\ln2\right)\) \(=2-\ln2\)
