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Câu a)
\(\int \frac{1}{\cos^4x}dx=\int \frac{\sin ^2x+\cos^2x}{\cos^4x}dx=\int \frac{\sin ^2x}{\cos^4x}dx+\int \frac{1}{\cos^2x}dx\)
Xét \(\int \frac{1}{\cos^2x}dx=\int d(\tan x)=\tan x+c\)
Xét \(\int \frac{\sin ^2x}{\cos^4x}dx=\int \frac{\tan ^2x}{\cos^2x}dx=\int \tan^2xd(\tan x)=\frac{\tan ^3x}{3}+c\)
Vậy :
\(\int \frac{1}{\cos ^4x}dx=\frac{\tan ^3x}{3}+\tan x+c\)
\(\Rightarrow \int ^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{dx}{\cos^4 x}=\)\(\left.\begin{matrix} \frac{\pi}{3}\\ \frac{\pi}{6}\end{matrix}\right|\left ( \frac{\tan ^3 x}{3}+\tan x+c \right )=\frac{44}{9\sqrt{3}}\)
Câu b)
\(\int \frac{(x+1)^2}{x^2+1}dx=\int \frac{x^2+1+2x}{x^2+1}dx=\int dx+\int \frac{2xdx}{x^2+1}\)
\(=x+c+\int \frac{d(x^2+1)}{x^2+1}=x+\ln (x^2+1)+c\)
Do đó:
\(\int ^{1}_{0}\frac{(x+1)^2}{x^2+1}dx=\left.\begin{matrix} 1\\ 0\end{matrix}\right|(x+\ln (x^2+1)+c)=\ln 2+1\)
Câu c)
\(\int \frac{x^2+2\ln x}{x}dx=\int xdx+2\int \frac{2\ln x}{x}dx\)
\(=\frac{x^2}{2}+c+2\int \ln xd(\ln x)\)
\(=\frac{x^2}{2}+c+\ln ^2x\)
\(\Rightarrow \int ^{2}_{1}\frac{x^2+2\ln x}{x}dx=\left.\begin{matrix} 2\\ 1\end{matrix}\right|\left ( \frac{x^2}{2}+\ln ^2x +c \right )=\frac{3}{2}+\ln ^22\)
Câu d)
\(\int^{2}_{1} \frac{x^2+3x+1}{x^2+x}dx=\int ^{2}_{1}dx+\int ^{2}_{1}\frac{2x+1}{x^2+x}dx\)
\(=\left.\begin{matrix} 2\\ 1\end{matrix}\right|x+\int ^{2}_{1}\frac{d(x^2+x)}{x^2+x}=1+\left.\begin{matrix} 2\\ 1\end{matrix}\right|\ln |x^2+x|=1+\ln 6-\ln 2\)
\(=1+\ln 3\)
\(\int\left(\frac{1}{x}-2x\right)dx=ln\left|x\right|-x^2+C\)
\(\int cos2xdx=\frac{1}{2}sin2x+C\)
\(\int\frac{1}{x^2-4x+4}dx=\int\frac{d\left(x-2\right)}{\left(x-2\right)^2}=-\frac{1}{\left(x-2\right)}+C=\frac{1}{2-x}+C\)
\(\int\limits^4_1\frac{1}{2\sqrt{x}}dx=\sqrt{x}|^4_1=\sqrt{4}-\sqrt{1}=1\)
\(I=\int\limits^1_0\left(2x+1\right)e^xdx\)
Đặt \(\left\{{}\begin{matrix}u=2x+1\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2dx\\v=e^x\end{matrix}\right.\)
\(\Rightarrow I=\left(2x+1\right)e^x|^1_0-\int\limits^1_02e^xdx=3e-1-2e^x|^1_0=e+3\)
\(\int\limits^{\dfrac{\pi}{4}}_{\dfrac{\pi}{8}}\dfrac{dx}{sin^2x.cos^2x}=\int\limits^{\dfrac{\pi}{4}}_{\dfrac{\pi}{8}}\dfrac{2d\left(2x\right)}{sin^22x}=-2cot2x|^{\dfrac{\pi}{4}}_{\dfrac{\pi}{8}}=...\)
\(\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\dfrac{cos2xdx}{sin^2x.cos^2x}=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\dfrac{cos^2x-sin^2x}{sin^2x.cos^2x}dx=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\left(\dfrac{1}{sin^2x}-\dfrac{1}{cos^2x}\right)dx=\left(-cotx-tanx\right)|^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\)
\(\int\limits^{\dfrac{\pi}{3}}_0\dfrac{cos3x}{cosx}dx=\int\limits^{\dfrac{\pi}{3}}_0\dfrac{4cos^3x-3cosx}{cosx}dx=\int\limits^{\dfrac{\pi}{3}}_0\left(4cos^2x-3\right)dx\)
\(=\int\limits^{\dfrac{\pi}{3}}_0\left(2cos2x-1\right)dx=\left(sin2x-x\right)|^{\dfrac{\pi}{3}}_0=...\)
Chúc bạn học tốt
Biến đổi: ʃ\(\int\dfrac{1dx}{cosx\dfrac{\sqrt{2}}{2}\left(cosx-sinx\right)}=\int\dfrac{\sqrt{2}dx}{cos^2x\left(1-tanx\right)}=\int\dfrac{\sqrt{2}d\left(tanx\right)}{1-tanx}=-\sqrt{2}\ln trituyetdoi\left(1-tanx\right)\)
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