1: \(\int \sin\left(\frac{\pi}{4} - x\right) dx\)

mà ta có công thức: \(\int \sin(ax + b) dx = -\frac{1}{a}\cos(ax + b) + C\) nên với a=-1; \(b=\frac{\pi}{4}\) thì

\(\int\sin\left(\frac{\pi}{4}-x\right)dx=-\frac{1}{-1}\cos\left(\frac{\pi}{4}-x\right)+C=\cos\left(\frac{\pi}{4}-x\right)+C\)

2: \(\int \frac{7}{\cos^2(3-x)} dx = \frac{7}{-1} \tan(3-x) = -7\tan(3-x)\)

\(\int 8\sin(9-3x) dx = 8 \cdot \left(-\frac{1}{-3}\right)\cos(9-3x) = \frac{8}{3}\cos(9-3x)\)

\(\int -\frac{1}{x} dx = -\ln\vert{}x\vert{}\)

\(\int \frac{6}{3-2x} dx = 6 \cdot \left(-\frac{1}{2}\right)\ln\vert{}3-2x\vert{} = -3\ln\vert{}3-2x\vert{}\)

\(\int \sqrt{x} dx = \int x^{\frac{1}{2}} dx = \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2}{3}x\sqrt{x}\)

\(\int \left( \frac{7}{\cos^2(3-x)} + 8\sin(9-3x) - \frac{1}{x} + \frac{6}{3-2x} + \sqrt{x} \right) dx\)

\(= -7\tan(3-x) + \frac{8}{3}\cos(9-3x) - \ln\vert{}x\vert{} - 3\ln\vert{}3-2x\vert{} + \frac{2}{3}x\sqrt{x} + C\)

3: \(\int \frac{7}{\cos^2 x} dx = 7\tan x\)

\(\int -\frac{8}{2x+1} dx = -8 \cdot \frac{1}{2} \ln\vert{}2x+1\vert{} = -4\ln\vert{}2x+1\vert{}\)

\(\int 9^{2x+1} dx = \frac{1}{2} \cdot \frac{9^{2x+1}}{\ln 9} = \frac{9^{2x+1}}{4\ln 3}\)

\(\int e^{5-2x} dx = -\frac{1}{2}e^{5-2x}\)

\(\int 8 dx = 8x\)

\(\int \left( \frac{7}{\cos^2 x} - \frac{8}{2x+1} + 9^{2x+1} + e^{5-2x} + 8 \right) dx\)

\(= 7\tan x - 4\ln\vert{}2x+1\vert{} + \frac{9^{2x+1}}{4\ln 3} - \frac{1}{2}e^{5-2x} + 8x + C\)

4: \(\int \frac{4}{x} dx = 4\ln\vert{}x\vert{}\)

\(\int -x^{-\frac{1}{2}} dx = -\frac{x^{\frac{1}{2}}}{\frac{1}{2}} = -2\sqrt{x}\)

\(\int 5x^4 dx = 5 \cdot \frac{x^5}{5} = x^5\)

\(\int -6x^6 dx = -6 \cdot \frac{x^7}{7} = -\frac{6}{7}x^7\)

\(\int\frac{3 - \sqrt{x} + 5x^5 - 6x^7 + 1}{x}dx\)

\(=\int\frac{4 - x^{\frac{1}{2}} + 5x^5 - 6x^7}{x}dx\)

\(= \int \left( \frac{4}{x} - x^{-\frac{1}{2}} + 5x^4 - 6x^6 \right) dx\)

\(= 4\ln\vert{}x\vert{} - 2\sqrt{x} + x^5 - \frac{6}{7}x^7 + C\)

a)

Ta có \(A=\int ^{\frac{\pi}{4}}_{0}\cos 2x\cos^2xdx=\frac{1}{4}\int ^{\frac{\pi}{4}}_{0}\cos 2x(\cos 2x+1)d(2x)\)

\(\Leftrightarrow A=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos x(\cos x+1)dx=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos xdx+\frac{1}{8}\int ^{\frac{\pi}{2}}_{0}(\cos 2x+1)dx\)

\(\Leftrightarrow A=\frac{1}{4}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin x+\frac{1}{16}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin 2x+\frac{1}{8}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|x=\frac{1}{4}+\frac{\pi}{16}\)

b)

\(B=\int ^{1}_{\frac{1}{2}}\frac{e^x}{e^{2x}-1}dx=\frac{1}{2}\int ^{1}_{\frac{1}{2}}\left ( \frac{1}{e^x-1}-\frac{1}{e^x+1} \right )d(e^x)\)

\(\Leftrightarrow B=\frac{1}{2}\left.\begin{matrix} 1\\ \frac{1}{2}\end{matrix}\right|\left | \frac{e^x-1}{e^x+1} \right |\approx 0.317\)

c)

Có \(C=\int ^{1}_{0}\frac{(x+2)\ln(x+1)}{(x+1)^2}d(x+1)\).

Đặt \(x+1=t\)

\(\Rightarrow C=\int ^{2}_{1}\frac{(t+1)\ln t}{t^2}dt=\int ^{2}_{1}\frac{\ln t}{t}dt+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)

\(=\int ^{2}_{1}\ln td(\ln t)+\int ^{2}_{1}\frac{\ln t}{t^2}dt=\frac{\ln ^22}{2}+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)

Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=\frac{dt}{t^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=\frac{-1}{t}\end{matrix}\right.\Rightarrow \int ^{2}_{1}\frac{\ln t}{t^2}dt=\left.\begin{matrix} 2\\ 1\end{matrix}\right|-\frac{\ln t+1}{t}=\frac{1}{2}-\frac{\ln 2 }{2}\)

\(\Rightarrow C=\frac{1}{2}-\frac{\ln 2}{2}+\frac{\ln ^22}{2}\)

\(=\int\left(6x^2-\dfrac{4}{x}+sin3x-cos4x+e^{2x+1}+9^{x-1}+\dfrac{1}{cos^2x}-\dfrac{1}{sin^2x}\right)dx\)

\(=2x^3-4ln\left|x\right|-\dfrac{1}{3}cos3x-\dfrac{1}{4}sin4x+\dfrac{1}{2}e^{2x+1}+\dfrac{9^{x-1}}{ln9}+tanx+cotx+C\)

Chọn B

Gọi r₁, r₂, r₃, r₄ là khoảng cánh từ điểm M đến bốn mặt của tứ diện.

Gọi S là diện tích một mặt của tứ diện 

Đường cao của tứ diện là .

Thể tích của tứ diện là .

Mặt khác, ta có:

Chọn C

\(\int\frac{1+sin2x+cos2x}{sinx+cosx}dx\)

\(=\int\frac{sin^2x+cos^2x+2sinxcosx+cos^2x-sin^2x}{sinx+cosx}dx\)

\(=\int\frac{\left(sinx+cosx\right)^2+\left(cosx-sinx\right)\left(cosx+sinx\right)}{sinx+cosx}dx\)

\(=\int\left(sinx+cosx+cosx-sinx\right)dx=\int2cosxdx=2sinx\)

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