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Lời giải:
Xét \(\int \frac{\tan ^2x-\cos ^2x}{\sin ^2x}dx=\int \frac{\tan ^2x}{\sin ^2x}dx-\int \frac{\cos ^2x}{\sin ^2x}dx\)
Có:
\(\int \frac{\tan ^2x}{\sin ^2x}dx=\int \frac{\sin ^2x}{\cos ^2x. \sin^2 x}dx=\int \frac{1}{\cos ^2x}dx\)
\(=\int d(\tan x)=\tan x+c\)
Và:
\(\int \frac{\cos ^2x}{\sin ^2x}dx=\int \frac{1-\sin ^2x}{\sin ^2x}dx=\int \frac{1}{\sin ^2x}dx-\int dx\)
\(=-\int d(\cot x)-x+c=-\cot x-x+c\)
Do đó:
\(\int \frac{\tan ^2x-\cos ^2x}{\sin ^2x}dx=\tan x+c-(-\cot x-x+c)=\tan x+\cot x+x+c\)
\(\Rightarrow \int ^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\tan ^2x-\cos ^2x}{\sin ^2x}dx=\frac{4\sqrt{3}}{3}+\frac{\pi}{3}-\frac{4\sqrt{3}}{3}-\frac{\pi}{6}=\frac{\pi}{6}\)
Ở tất cả các dạng bài như thế này em chỉ cần ghi nhớ công thức:
\(d(u(x))=u'(x)dx\)
Câu 1)
Ta có \(I_1=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} e^{\sin x}\cos xdx=\int _{\frac{\pi}{4}}^{\frac{\pi}{2}}e^{\sin x}d(\sin x)\)
Đặt \(\sin x=t\Rightarrow I_1=\int ^{1}_{\frac{\sqrt{2}}{2}}e^tdt=\left.\begin{matrix} 1\\ \frac{\sqrt{2}}{2}\end{matrix}\right|e^t=e-e^{\frac{\sqrt{2}}{2}}\)
Câu 2)
\(I_2=\int ^{\frac{\pi}{2}}_{\frac{\pi}{4}}e^{2\cos x+1}\sin xdx=\frac{-1}{2}\int ^\frac{\pi}{2}_{\frac{\pi}{4}}e^{2\cos x+1}d(2\cos x+1)\)
Đặt \(2\cos x+1=t\Rightarrow I_2=\frac{-1}{2}\int ^{1}_{1+\sqrt{2}}e^tdt\)
\(=\frac{-1}{2}.\left.\begin{matrix} 1\\ 1+\sqrt{2}\end{matrix}\right|e^t=\frac{-1}{2}(e-e^{1+\sqrt{2}})\)
Câu 3:
Có \(I_3=\int ^{e}_{1}\frac{e^{2\ln x+1}}{x}dx=\int ^{e}_{1}e^{2\ln x+1}d(\ln x)\)
\(=\frac{1}{2}\int ^{e}_{1}e^{2\ln x+1}d(2\ln x+1)\)
Đặt \(2\ln x+1=t\Rightarrow I_3=\frac{1}{2}\int ^{3}_{1}e^tdt=\frac{1}{2}.\left.\begin{matrix} 3\\ 1\end{matrix}\right|e^t=\frac{1}{2}(e^3-e)\)
Câu 4:
\(I_4=\int ^{1}_{0}xe^{x^2+2}dx=\frac{1}{2}\int ^{1}_{0}e^{x^2+2}d(x^2+2)\)
Đặt \(x^2+2=t\Rightarrow I_4=\frac{1}{2}\int ^{3}_{2}e^tdt=\frac{1}{2}.\left.\begin{matrix} 3\\ 2\end{matrix}\right|e^t=\frac{1}{2}(e^3-e^2)\)
1)
Ta có:
\(\int (2-\cot ^2x)dx=\int (2-\frac{\cos ^2x}{\sin ^2x})dx\)
\(=\int (2-\frac{1-\sin ^2x}{\sin ^2x})dx=\int (3-\frac{1}{\sin ^2x})dx=3\int dx-\int \frac{dx}{\sin ^2x}\)
\(=3x+\int d(\cot x)=3x+\cot x+c\)
\(\Rightarrow \int ^{\frac{\pi}{2}}_{\frac{\pi}{3}}(2-\cot ^2x)dx=\left.\begin{matrix} \frac{\pi}{2}\\ \frac{\pi}{3}\end{matrix}\right|(3x+\cot x+c)=\frac{\pi}{2}-\frac{\sqrt{3}}{3}\)
3)
Xét \(\int (2\tan x-3\cot x)^2dx\)
\(=\int (4\tan ^2x+9\cot ^2x-12)dx\)
\(=\int (\frac{4\sin ^2x}{\cos ^2x}+\frac{9\cos ^2x}{\sin ^2x}-12)dx\)
\(=\int (\frac{4(1-\cos ^2x)}{\cos ^2x}+\frac{9(1-\sin ^2x)}{\sin ^2x}-12)dx\)
\(=\int (\frac{4}{\cos ^2x}+\frac{9}{\sin ^2x}-25)dx\)
\(=4\int d(\tan x)-9\int d(\cot x)-25\int dx\)
\(=4\tan x-9\cot x-25x+c\)
Do đó:
\(\int ^{\frac{\pi}{3}}_{\frac{\pi}{6}}(2\tan x-3\cot x)^2dx=\left.\begin{matrix} \frac{\pi}{3}\\ \frac{\pi}{6}\end{matrix}\right|(4\tan x-9\cot x-25x+c)=\frac{26\sqrt{3}}{3}-\frac{25\pi}{6}\)
2)
Xét \(\int (\tan x+\cot x)^2dx=\int (\tan ^2x+\cot ^2x+2)dx\)
\(=\int (\frac{\sin ^2x}{\cos^2 x}+\frac{\cos ^2x}{\sin ^2x}+2)dx\)
\(=\int (\frac{1-\cos ^2x}{\cos ^2x}+\frac{1-\sin ^2x}{\sin ^2x}+2)dx\)
\(=\int (\frac{1}{\cos ^2x}+\frac{1}{\sin ^2x})dx\)
\(=\int d(\tan x)-\int d(\cot x)=\tan x-\cot x+c\)
Do đó:
\(\int ^{\frac{\pi}{3}}_{\frac{\pi}{6}}(\tan x+\cot x)^2dx=\left.\begin{matrix} \frac{\pi}{3}\\ \frac{\pi}{6}\end{matrix}\right|(\tan x-\cot x+c)=2\sqrt{3}-\frac{2\sqrt{3}}{3}\)
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)
\(\int^{2}_{1} \frac{x^2+3x+1}{x^2+x}dx=\int ^{2}_{1}dx+\int ^{2}_{1}\frac{2x+1}{x^2+x}dx\)
\(=\left.\begin{matrix} 2\\ 1\end{matrix}\right|x+\int ^{2}_{1}\frac{d(x^2+x)}{x^2+x}=1+\left.\begin{matrix} 2\\ 1\end{matrix}\right|\ln |x^2+x|=1+\ln 6-\ln 2\)
\(=1+\ln 3\)
Câu nào mình biết thì mình làm nha.
1) Đổi thành \(\dfrac{y^4}{4}+y^3-2y\) rồi thế số.KQ là \(\dfrac{-3}{4}\)
2) Biến đổi thành \(\dfrac{t^2}{2}+2\sqrt{t}+\dfrac{1}{t}\) và thế số.KQ là \(\dfrac{35}{4}\)
3) Biến đổi thành 2sinx + cos(2x)/2 và thế số.KQ là 1
\(\int\limits^{\dfrac{\pi}{4}}_{\dfrac{\pi}{8}}\dfrac{dx}{sin^2x.cos^2x}=\int\limits^{\dfrac{\pi}{4}}_{\dfrac{\pi}{8}}\dfrac{2d\left(2x\right)}{sin^22x}=-2cot2x|^{\dfrac{\pi}{4}}_{\dfrac{\pi}{8}}=...\)
\(\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\dfrac{cos2xdx}{sin^2x.cos^2x}=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\dfrac{cos^2x-sin^2x}{sin^2x.cos^2x}dx=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\left(\dfrac{1}{sin^2x}-\dfrac{1}{cos^2x}\right)dx=\left(-cotx-tanx\right)|^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\)
\(\int\limits^{\dfrac{\pi}{3}}_0\dfrac{cos3x}{cosx}dx=\int\limits^{\dfrac{\pi}{3}}_0\dfrac{4cos^3x-3cosx}{cosx}dx=\int\limits^{\dfrac{\pi}{3}}_0\left(4cos^2x-3\right)dx\)
\(=\int\limits^{\dfrac{\pi}{3}}_0\left(2cos2x-1\right)dx=\left(sin2x-x\right)|^{\dfrac{\pi}{3}}_0=...\)