Tìm nguyên hàm F(x) của hàm số \(f\left(x\right)=\frac{1}{x}\) trên khoảng \(\left(0;+\infty\right)\), biết \(F\left(e\right)=2e\). \(F\left(x\right)=1+2e-\ln x\) \(F\left(x\right)=\ln x+2e-1\) \(F\left(x\right)=-\frac{1}{x^2}+2e+\frac{1}{e^2}\) \(F\left(x\right)=\ln\left|x+2e-1\right|\) Hướng dẫn giải: \(F\left(x\right)=\int\frac{1}{x}\text{dx}=\ln\left|x\right|+C\) \(=\ln x+C,x>0\) Vì \(F\left(e\right)=2e\) nên suy ra: \(\ln e+C=2e\) \(\Rightarrow C=2e-\ln e=2e-1\) Vậy \(F\left(x\right)=\ln x+2e-1\)
Tìm nguyên hàm \(f\left(x\right)\) của hàm số \(\cos2x\), biết \(f\left(\frac{\pi}{2}\right)=2\pi\). \(f\left(x\right)=\sin x+2\pi\) \(f\left(x\right)=x+\sin2x+\frac{3\pi}{2}\) \(f\left(x\right)=2x+\pi\) \(f\left(x\right)=\frac{1}{2}\sin2x+2\pi\) Hướng dẫn giải: \(f\left(x\right)=\int\cos2x\text{dx}=\frac{1}{2}\int\cos2x\text{d}\left(2x\right)=\frac{1}{2}\sin2x+C\) Vì \(f\left(\frac{\pi}{2}\right)=2\pi\) nên \(\frac{1}{2}\sin\left(2.\frac{\pi}{2}\right)+C=2\pi\) \(\Rightarrow C=2\pi\) Vậy \(f\left(x\right)=\frac{1}{2}\sin2x+2\pi\)
Tìm nguyên hàm \(F\left(x\right)\) của hàm số \(f\left(x\right)=\cos3x\cos x\), bieets ddoof thij \(y=F\left(x\right)\) đi qua gốc tọa độ. \(F\left(x\right)=\frac{\sin4x}{4}+\frac{\sin2x}{2}\) \(F\left(x\right)=\frac{\sin4x}{8}+\frac{\sin2x}{4}\) \(F\left(x\right)=\frac{\cos4x}{8}+\frac{\cos2x}{4}\) \(F\left(x\right)=\frac{\sin8x}{8}+\frac{\sin4x}{4}\) Hướng dẫn giải: \(f\left(x\right)=\cos3x\cos x=\frac{1}{2}\left[\cos\left(3x+x\right)+\cos\left(3x-x\right)\right]=\frac{1}{2}\left(\cos4x+\cos2x\right)\) \(F\left(x\right)=\frac{1}{2}\int\left(\cos4x+\cos2x\right)\text{dx}=\frac{1}{2}\left[\frac{1}{4}\int\cos4xd\left(4x\right)+\frac{1}{2}\int\cos2xd\left(2x\right)\right]\) \(=\frac{1}{2}\left[\frac{1}{4}\sin4x+\frac{1}{2}\sin2x\right]=\frac{\sin4x}{8}+\frac{\sin2x}{4}\)
Tìm hàm số F(x) trên khoảng \(\left(0;\frac{\pi}{2}\right)\) thỏa mãn các điều kiện sau: Có số a, b để: \(F'\left(x\right)=\frac{a\sin^2x\cos^2x+b\sqrt{3}}{\sin^2x\cos^2x},F\left(\frac{\pi}{6}\right)=\frac{\pi}{2},F\left(\frac{\pi}{4}\right)=\frac{\pi}{4},F\left(\frac{\pi}{3}\right)=\pi\) \(F\left(x\right)=9x-2\pi\) \(F\left(x\right)=x+\frac{\pi}{\sqrt{3}}\left(\tan x-\cot x\right)-\frac{\pi}{12}\) \(F\left(x\right)=x+\frac{\pi}{\sqrt{3}}\left(\tan x-\cot x\right)\) \(F\left(x\right)=x+\frac{\pi}{\sqrt{3}}\left(\tan x-\cot x\right)+\frac{\pi}{6}\) Hướng dẫn giải: \(F\left(x\right)=\int\frac{a\sin^2x\cos^2x+b\sqrt{3}}{\sin^2x\cos^2x}\text{dx}=\int\left(a+\frac{b\sqrt{3}}{\sin^2x\cos^2x}\right)\text{dx}\) \(=\int\left[a+\left(\frac{1}{\sin^2x}+\frac{1}{\cos^2x}\right)b\sqrt{3}\right]\text{dx}\) \(=ax+\left(-\cot x+\tan x\right)b\sqrt{3}+C\) Để thỏa mãn các điều kiện: \(F\left(\frac{\pi}{6}\right)=\frac{\pi}{2},F\left(\frac{\pi}{4}\right)=\frac{\pi}{4},F\left(\frac{\pi}{3}\right)=\pi\) ta có: \(\left\{\begin{matrix}a.\frac{\pi}{6}+\left(-\cot\frac{\pi}{6}+\tan\frac{\pi}{6}\right)b\sqrt{3}+C=\frac{\pi}{2}\\a\frac{\pi}{4}+\left(-\cot\frac{\pi}{4}+\tan\frac{\pi}{4}\right)b\sqrt{3}+C=\frac{\pi}{4}\\a\frac{\pi}{3}+\left(-\cot\frac{\pi}{3}+\tan\frac{\pi}{3}\right)b\sqrt{3}+C=\pi\end{matrix}\right.\) Giải hệ theo a, b, C ta được: \(a=1,b=\frac{\pi}{3},C=0\). Vậy \(F\left(x\right)=x+\frac{\pi}{\sqrt{3}}\left(\tan x-\cot x\right)\)
Cho f(x) là một hàm số có đạo hàm f\(f'\left(x\right)\) liên tục trên đoạn \(\left[0;\frac{\pi}{2}\right]\) thỏa mãn điều kiện \(f\left(0\right)=\frac{\pi}{2}\). Biết \(\int\limits^{\frac{\pi}{2}}_0f'\left(x\right)\text{dx}=2\pi\). Tính \(f\left(\frac{\pi}{2}\right)\). \(f\left(\frac{\pi}{2}\right)=\frac{\pi}{2}\) \(f\left(\frac{\pi}{2}\right)=0\) \(f\left(\frac{\pi}{2}\right)=\frac{3\pi}{2}\) \(f\left(\frac{\pi}{2}\right)=\frac{5\pi}{2}\) Hướng dẫn giải: Ta có: \(\int\limits^{\dfrac{\pi}{2}}_0f'\left(x\right)\text{dx}=f\left(\dfrac{\pi}{2}\right)-f\left(0\right)\) Vậy: \(2\pi=f\left(\dfrac{\pi}{2}\right)-f\left(0\right)=f\left(\dfrac{\pi}{2}\right)-\dfrac{\pi}{2}\) \(\Rightarrow f\left(\dfrac{\pi}{2}\right)=2\pi+\dfrac{\pi}{2}=\dfrac{5\pi}{2}\)
Cho hàm số f(x) có \(\int\limits^9_0f\left(x\right)\text{dx}=9\). Tính \(\int\limits^3_0f\left(3x\right)\text{dx}\). \(\int\limits^3_0f\left(3x\right)\text{dx}=3\) \(\int\limits^3_0f\left(3x\right)\text{dx}=27\) \(\int\limits^3_0f\left(3x\right)\text{dx}=-3\) \(\int\limits^3_0f\left(3x\right)\text{dx}=1\) Hướng dẫn giải: Đặt \(x=3t\) khi đó \(dx=3\text{dt}\), \(x|^9_0\Rightarrow t|^3_0\) \(\int\limits^9_0f\left(x\right)\text{dx}=\int\limits^3_0f\left(3t\right)\text{3dt}=3\int\limits^3_0f\left(3t\right)\text{d}t=3\int\limits^3_0f\left(3x\right)\text{dx}\) Suy ra \(\int\limits^3_0f\left(3x\right)\text{dx}=\dfrac{1}{3}\int\limits^9_0f\left(x\right)\text{dx}=\dfrac{1}{3}.9=3\)
Tính \(S=e^{\int\limits^5_1\dfrac{\text{dx}}{2x-1}}\). 3 8 18 25 Hướng dẫn giải: Ta có: \(\int\limits^5_1\dfrac{dx}{2x-1}=\dfrac{1}{2}\int\limits^5_1\dfrac{d\left(2x-1\right)}{2x-1}=\ln\left|2x-1\right||^5_1=\ln9-\ln3=\ln3^2-\ln3=2\ln3-\ln3=\ln3\) Suy ra \(S=e^{\ln3}=3\).
Trong các hàm số \(f\left(t\right)\) sau, hàm số nào thỏa mãn \(\int\limits^{\dfrac{\pi}{4}}_0\left(1-\tan x\right)^4\dfrac{1}{\cos^2x}\text{dx}=\int\limits^1_0f\left(t\right)\text{dt}\) \(f\left(t\right)=t^4\) \(f\left(t\right)=t^2\) \(f\left(t\right)=\left(1-t\right)^2\) \(f\left(t\right)=\left(t-1\right)^3\) Hướng dẫn giải: Đặt \(t=1-\tan x\) thì \(\text{dt}=-\dfrac{1}{\cos^2x}\text{dx}\) và \(x|^{\dfrac{\pi}{4}}_0\Rightarrow t|^0_1\), ta có: \(\int\limits^{\dfrac{\pi}{4}}_0\left(1-\tan x\right)^4\dfrac{1}{\cos^2x}\text{dx}=-\int\limits^0_1t^4dt=\int\limits^1_0t^4\text{dt}\) Đối sánh với hàm f(t) trong đầu bài thì \(f\left(t\right)=t^4\).
Đặt \(t=1+\sqrt{x-1}\). Hãy tìm hàm số \(f\left(t\right)\) trong các hàm số sau để \(\int\limits^2_1\dfrac{x\text{dx}}{1+\sqrt{x-1}}=\int\limits^2_1f\left(t\right)\text{dt}\). \(f\left(t\right)=\dfrac{2t^3+2t}{t+1}\) \(f\left(t\right)=2t^2-6t+8-\dfrac{4}{t}\) \(f\left(t\right)=2t\sqrt{t}-1\) \(f\left(t\right)=3t-\ln\left|t\right|\) Hướng dẫn giải: \(t=1+\sqrt{x-1}\) \(\Rightarrow x=\left(t-1\right)^2+1\) và \(\text{dt}=\dfrac{1}{2\sqrt{x-1}}\text{dx}\) \(\Rightarrow\text{dx}=2\sqrt{x-1}\text{dt}=2\left(t-1\right)\text{dt}\) Đổi cận \(x|^2_1\Rightarrow t|^2_1\) Suy ra: \(\int\limits^2_1\dfrac{x\text{dx}}{1+\sqrt{x-1}}=\int\limits^2_1\dfrac{\left(t-1\right)^2+1}{t}2\left(t-1\right)\text{dt}=\int\limits^2_1\left(2t^2-6t+8-\dfrac{4}{t}\right)\text{dt}\) Vậy \(f\left(t\right)=2t^2-6t+8-\dfrac{4}{t}\).
Đặt \(I=\int\limits^{\ln5}_{\ln3}\dfrac{\text{dx}}{e^x+2e^{-x}-3}\) và \(t=e^x\). Trong các khẳng định sau, khẳng định nào sai? \(I=\ln3-\ln2\) \(\dfrac{1}{e^x+2e^{-x}-3}=\dfrac{t}{\left(t-1\right)\left(t-2\right)}\) \(I=\int\limits^5_3\left(\dfrac{1}{t-2}-\dfrac{1}{t-1}\right)\text{dt}\) \(\int\left(\dfrac{1}{t-2}-\dfrac{1}{t-1}\right)\text{dt}=\ln\dfrac{t-2}{t-1}+C\) Hướng dẫn giải: Đặt \(t=e^x\) thì: \(\dfrac{1}{e^x+2e^{-x}-3}=\dfrac{1}{t+2.\dfrac{1}{t}-3}=\dfrac{t}{t^2-3t+2}=\dfrac{t}{\left(t-1\right)\left(t-2\right)}\) Ta lại có: \(\dfrac{1}{\left(t-1\right)\left(t-2\right)}=\dfrac{1}{t-2}-\dfrac{1}{t-1}\) nên \(I=\int\limits^{\ln5}_{\ln3}\dfrac{\text{dx}}{e^x+2e^{-x}-3}=\int\limits^5_3\left(\dfrac{1}{t-2}-\dfrac{1}{t-1}\right)\text{dt}\) Vì \(\int\left(\dfrac{1}{t-2}-\dfrac{1}{t-1}\right)\text{dt}=\int\dfrac{d\left(t-2\right)}{t-2}-\int\dfrac{d\left(t-1\right)}{t-1}=\ln\left|t-2\right|-\ln\left|t-1\right|+C=\ln\left|\dfrac{t-2}{t-1}\right|+C\) nên \(I=\ln\left|\dfrac{t-2}{t-1}\right||^5_3=\ln\left|\dfrac{5-2}{5-1}\right|-\ln\left|\dfrac{3-2}{3-1}\right|=\ln\dfrac{3}{4}-\ln\dfrac{1}{2}\) \(=\ln3-\ln4-\ln1+\ln2=\ln3-2\ln2+\ln2=\ln3-\ln2\) Chú ý: \(\int\left(\dfrac{1}{t-2}-\dfrac{1}{t-1}\right)\text{dt}=\ln\left|\dfrac{t-2}{t-1}\right|+C\)
