\(=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{3\left(1-cos^2x\right)-4cos^2x}{cos^2x}dx=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{3-7cos^2x}{cos^2x}dx\)

\(=\int\limits^{\dfrac{\pi}{4}}_0\left(\dfrac{3}{cos^2x}-7\right)dx=\left(3tanx-7x\right)|^{\dfrac{\pi}{4}}_0=...\)

\(=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{2cos^2x-1-3\left(1-cos^2x\right)}{cos^2x}dx=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{5cos^2x-4}{cos^2x}dx\)

\(=\int\limits^{\dfrac{\pi}{4}}_0\left(5-\dfrac{4}{cos^2x}\right)dx=\left(5x-4tanx\right)|^{\dfrac{\pi}{4}}_0=...\)

1)

\(I=\int\left(cos^2x-cos^2x\cdot sin^3x\right)dx\\ =\int cos^2x\cdot dx-\int cos^2x\cdot sin^3x\cdot dx\\ =\frac{1}{2}\int\left(cos2x+1\right)dx+\int cos^2x\left(1-cos^2x\right)d\left(cosx\right)\\ =\frac{1}{4}sin2x+\frac{1}{2}+\frac{cos^3x}{3}-\frac{cos^5x}{5}+C\)

....

2) Xét riêng mẫu số:

\(sin2x+2\left(1+sinx+cosx\right)\\ =\left(sin2x+1\right)+2\left(sinx+cosx\right)+1\\ =\left(sinx+cosx\right)^2+2\left(sinx+cosx\right)+1\\ =\left(sinx+cosx+1\right)^2\\ =\left[\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1\right]^2\)

Khi đó:

\(I_2=\int\frac{sin\left(x-\frac{\pi}{4}\right)}{\left[\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1\right]^2}dx\\ =-\frac{1}{\sqrt{2}}\int\frac{d\left[\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1\right]}{\left[\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1\right]^2}\\ =\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1}+C=\frac{1}{2cos\left(x-\frac{\pi}{4}\right)+1}\)

...

Câu 1)

Đặt \(\left\{\begin{matrix} u=\ln ^2x\\ dv=\frac{1}{x^2}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{2\ln x}{x}\\ v=\frac{-1}{x}\end{matrix}\right.\)

\(\int \left ( \frac{\ln}{x} \right )^2dx=\frac{-\ln^2x}{x}+2\int \frac{\ln x}{x^2}dx\)

Đặt \(\left\{\begin{matrix} t=\ln x\\ dk=\frac{1}{x^2}dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} dt=\frac{1}{x}dx\\ k=-\frac{1}{x}\end{matrix}\right.\Rightarrow \int \frac{\ln x}{x^2}dx=-\frac{\ln x}{x}+\int \frac{1}{x^2}dx=\frac{-\ln x}{x}-\frac{1}{x}\)

\(\Rightarrow I=\left.\begin{matrix} e\\ 1\end{matrix}\right|\left(\frac{-\ln^2 x}{x}-\frac{2\ln x}{x}-\frac{2}{x}\right)=2-\frac{5}{e}\)

Câu 2)

\(I=\int ^{\frac{\pi}{4}}_{0}\frac{x}{1+\cos 2x}dx=\frac{1}{2}\int ^{\frac{\pi}{4}}_{0}\frac{x}{\cos^2x}dx\)

Đặt \(\left\{\begin{matrix} u=x\\ dv=\frac{dx}{\cos^2x}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\tan x\end{matrix}\right.\Rightarrow I=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\frac{x\tan x}{2}-\frac{1}{2}\int^{\frac{\pi}{4}}_{0} \tan xdx\)

\(=\frac{\pi}{8}+\frac{1}{2}\int ^{\frac{\pi}{4}}_{0}\frac{d(\cos x)}{\cos x}=\frac{\pi}{8}+\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\frac{\ln |\cos x|}{2}=\frac{\pi}{8}+\frac{\ln\frac{\sqrt{2}}{2}}{2}\)

Câu 3)

Đặt \(\left\{\begin{matrix} u=\ln (\cos x)\\ dv=\frac{dx}{\cos^2x}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{-\sin x}{\cos x}dx=-\tan xdx\\ v=\tan x\end{matrix}\right.\)

\(\Rightarrow I=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\tan x\ln (\cos x)+\int ^{\frac{\pi}{4}}_{0}\tan^2xdx=\ln \frac{\sqrt{2}}{2}+\int ^{\frac{\pi}{4}}_{0}(\frac{1}{\cos^2x}-1)dx\)

\(=\ln\frac{\sqrt{2}}{2}+\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|(\tan x-x)=\ln \frac{\sqrt{2}}{2}-\frac{\pi}{4}+1\)

Nhìn đề dữ dội y hệt cr của tui z :( Để làm từ từ 

Lập bảng xét dấu cho \(\left|x^2-1\right|\) trên đoạn \(\left[-2;2\right]\)

\(\left(-2;-1\right):+\)

\(\left(-1;1\right):-\)

\(\left(1;2\right):+\)

\(\Rightarrow I=\int\limits^{-1}_{-2}\left|x^2-1\right|dx+\int\limits^1_{-1}\left|x^2-1\right|dx+\int\limits^2_1\left|x^2-1\right|dx\)

\(=\int\limits^{-1}_{-2}\left(x^2-1\right)dx-\int\limits^1_{-1}\left(x^2-1\right)dx+\int\limits^2_1\left(x^2-1\right)dx\)

\(=\left(\dfrac{x^3}{3}-x\right)|^{-1}_{-2}-\left(\dfrac{x^3}{3}-x\right)|^1_{-1}+\left(\dfrac{x^3}{3}-x\right)|^2_1\)

Bạn tự thay cận vô tính nhé :), hiện mình ko cầm theo máy tính 

2/ \(I=\int\limits^e_1x^{\dfrac{1}{2}}.lnx.dx\)

\(\left\{{}\begin{matrix}u=lnx\\dv=x^{\dfrac{1}{2}}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{2}{3}.x^{\dfrac{3}{2}}\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}\int\limits^e_1x^{\dfrac{1}{2}}.dx\)

\(=\dfrac{2}{3}.x^{\dfrac{3}{2}}.lnx|^e_1-\dfrac{2}{3}.\dfrac{2}{3}.x^{\dfrac{3}{2}}|^e_1=...\)

3/ \(I=\int\limits^{\dfrac{\pi}{2}}_0e^{\sin x}.\cos x.dx+\int\limits^{\dfrac{\pi}{2}}_0\cos^2x.dx\)

Xét \(A=\int\limits^{\dfrac{\pi}{2}}_0e^{\sin x}.\cos x.dx\)

\(t=\sin x\Rightarrow dt=\cos x.dx\Rightarrow A=\int\limits^{\dfrac{\pi}{2}}_0e^t.dt=e^{\sin x}|^{\dfrac{\pi}{2}}_0\)

Xét \(B=\int\limits^{\dfrac{\pi}{2}}_0\cos^2x.dx\)

\(=\int\limits^{\dfrac{\pi}{2}}_0\dfrac{1+\cos2x}{2}.dx=\dfrac{1}{2}.\int\limits^{\dfrac{\pi}{2}}_0dx+\dfrac{1}{2}\int\limits^{\dfrac{\pi}{2}}_0\cos2x.dx\)

\(=\dfrac{1}{2}x|^{\dfrac{\pi}{2}}_0+\dfrac{1}{2}.\dfrac{1}{2}\sin2x|^{\dfrac{\pi}{2}}_0\)

I=A+B=...

d.

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^4x\)

\(tan^4x-3tan^2x-4tanx-3=0\)

\(\Leftrightarrow\left(tan^2x+tanx+1\right)\left(tan^2x-tanx-3\right)=0\)

\(\Leftrightarrow tan^2x-tanx-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1-\sqrt{13}}{2}\\tanx=\frac{1+\sqrt{13}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=arctan\left(\frac{1-\sqrt{13}}{2}\right)+k\pi\\x=arctan\left(\frac{1+\sqrt{13}}{2}\right)+k\pi\end{matrix}\right.\)

mọi người giúp hộ mình nhanh với

a.

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(2tan^3x+4=3tanx\left(1+tan^2x\right)\)

\(\Leftrightarrow2tan^3x+4=3tanx+3tan^3x\)

\(\Leftrightarrow tan^3x+3tanx-4=0\)

\(\Leftrightarrow\left(tanx-1\right)\left(tan^2x+tanx+4\right)=0\)

\(\Leftrightarrow tanx=1\Rightarrow x=\frac{\pi}{4}+k\pi\)

\(y'=3cosx-4sinx-\frac{1}{cos^2x}\)

\(\Rightarrow y'\left(\frac{\pi}{6}\right)=3cos\left(\frac{\pi}{6}\right)-4sin\left(\frac{\pi}{6}\right)-\frac{1}{cos^2\left(\frac{\pi}{6}\right)}=\frac{-20+9\sqrt{3}}{6}\)

b/ \(y'=-8x^3+\frac{3}{x^4}-\frac{1}{x^2}\)