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Câu 1)
Ta có \(I=\int ^{1}_{0}\frac{dx}{\sqrt{3+2x-x^2}}=\int ^{1}_{0}\frac{dx}{4-(x-1)^2}\).
Đặt \(x-1=2\cos t\Rightarrow \sqrt{4-(x-1)^2}=\sqrt{4-4\cos^2t}=2|\sin t|\)
Khi đó:
\(I=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}\frac{d(2\cos t+1)}{2\sin t}=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}\frac{2\sin tdt}{2\sin t}=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}dt=\left.\begin{matrix} \frac{2\pi}{3}\\ \frac{\pi}{2}\end{matrix}\right|t=\frac{\pi}{6}\)
Câu 3)
\(K=\int ^{3}_{2}\ln (x^3-3x+2)dx=\int ^{3}_{2}\ln [(x+2)(x-1)^2]dx\)
\(=\int ^{3}_{2}\ln (x+2)d(x+2)+2\int ^{3}_{2}\ln (x-1)d(x-1)\)
Xét \(\int \ln tdt\): Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln t dt=t\ln t-t\)
\(\Rightarrow K=\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x+2)[\ln (x+2)-1]+2\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x-1)[\ln (x-1)-1]\)
\(=5\ln 5-4\ln 4-1+4\ln 2-2=5\ln 5-4\ln 2-3\)
Bài 2)
\(J=\int ^{1}_{0}x\ln (2x+1)dx\). Đặt \(\left\{\begin{matrix} u=\ln (2x+1)\\ dv=xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{2dx}{2x+1}\\ v=\frac{x^2}{2}\end{matrix}\right.\)
Khi đó:
\(J=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2\ln (2x+1)}{2}-\int ^{1}_{0}\frac{x^2}{2x+1}dx\)\(=\frac{\ln 3}{2}-\frac{1}{4}\int ^{1}_{0}(2x-1+\frac{1}{2x+1})dx\)
\(=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2-x}{4}-\frac{1}{8}\int ^{1}_{0}\frac{d(2x+1)}{2x+1}=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{\ln (2x+1)}{8}\)
\(=\frac{\ln 3}{2}-\frac{\ln 3}{8}=\frac{3\ln 3}{8}\)
Bạn xem lại xem có type thiếu đề không? \((x+\frac{\pi}{6})\) có sin hay cos, tan ở phía trước không?
\(=\frac{1}{2}\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\ln\left(\tan x\right)d\left[\ln\left(\tan x\right)\right]=\frac{1}{4}\left[\ln^2\left(\tan x\right)\right]|^{\frac{\pi}{3}}_{\frac{\pi}{4}}=\frac{1}{4}\left(\ln^2\sqrt{3}-0\right)=\frac{1}{16}\ln^23\)
Đặt \(t=\tan x\Rightarrow\begin{cases}dt=\frac{dt}{\cos^2}=\left(1+t^2\right)dx\rightarrow dx=\frac{dt}{1+t^2}\\x=\frac{\pi}{4}\rightarrow t=1;x=\frac{\pi}{3}\rightarrow t=\sqrt{3}\end{cases}\)
Khi đó : \(I=\int\limits^{\sqrt{3}}_1\frac{\ln t}{\frac{2t}{1+t^2}}.\frac{dt}{1+t^2}=\frac{1}{2}\int\limits^{\sqrt{3}}_1\frac{\ln t}{t}dt=\frac{1}{2}J\left(1\right)\)
\(J=\int\limits^{\sqrt{3}}_1\frac{\ln t}{t}dt=\int\limits^{\sqrt{3}}_1\ln.d\left(\ln t\right)=\frac{1}{2}\ln^2t|^{\sqrt{3}}_1=\frac{1}{2}\left(\ln^2\sqrt{3}-0\right)=\frac{1}{8}\ln^23\)
Thay vào (1) ta có : \(I=\frac{1}{16}\ln^23\)
\(I=\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{\cos2x+3\cos x+2}dx=\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{2\cos^2x+3\cos x+1}dx\)
Đặt \(\cos x=t\Rightarrow dt=-\sin dx\)
Với \(x=0\Rightarrow t=1\)
Với \(x=\frac{\pi}{2}\Rightarrow t=0\)
\(I=\int\limits^1_0\frac{dt}{2t^2+3t+1}=\int\limits^1_0\frac{dt}{\left(2t+1\right)\left(t+1\right)}=2\int\limits^1_0\left(\frac{1}{2t+1}+\frac{1}{2t+1}\right)dt\)
\(=\left(\ln\frac{2t+1}{2t+1}\right)|^1_0=\ln\frac{3}{2}\)
Ta có \(I=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{\ln2.\ln\left(2\tan x\right)}{\sin2x.\ln\left(2\tan x\right)}dx=\ln2\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x.\ln\left(2\tan x\right)}+\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x}\)
Tính \(\ln2\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x.\ln\left(2\tan x\right)}=\frac{\ln2}{2}\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{d\left[\ln\left(2\tan x\right)\right]}{\ln2\left(2\tan x\right)}=\frac{\ln2}{2}\left[\ln\left(\ln\left(2\tan x\right)\right)\right]|^{\frac{\pi}{3}}_{\frac{\pi}{4}}=\frac{\ln2}{2}.\ln\left(\frac{\ln2\sqrt{3}}{\ln2}\right)\)
Tính \(\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x}=\frac{1}{2}\ln\left(\tan x\right)|^{\frac{\pi}{3}}_{\frac{\pi}{4}}=\frac{1}{2}\ln\sqrt{3}\)
Vậy \(I=\frac{\ln2}{2}\ln\left(\frac{\ln2\sqrt{3}}{\ln2}\right)+\frac{1}{2}\ln\sqrt{3}\)
\(\int\frac{1+sin2x+cos2x}{sinx+cosx}dx\)
\(=\int\frac{sin^2x+cos^2x+2sinxcosx+cos^2x-sin^2x}{sinx+cosx}dx\)
\(=\int\frac{\left(sinx+cosx\right)^2+\left(cosx-sinx\right)\left(cosx+sinx\right)}{sinx+cosx}dx\)
\(=\int\left(sinx+cosx+cosx-sinx\right)dx=\int2cosxdx=2sinx\)
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