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

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

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

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

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

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

b: \(cosx+3\cdot\sin\left(\frac{x}{2}\right)-2=0\)

=>\(cos\left(2\cdot\frac{x}{2}\right)+3\cdot\sin\left(\frac{x}{2}\right)-2=0\)

=>\(1-2\cdot\sin^2\left(\frac{x}{2}\right)+3\cdot\sin\left(\frac{x}{2}\right)-2=0\)

=>\(-2\cdot\sin^2\left(\frac{x}{2}\right)+3\cdot\sin\left(\frac{x}{2}\right)-1=0\)

=>\(2\cdot\sin^2\left(\frac{x}{2}\right)-3\cdot\sin\left(\frac{x}{2}\right)+1=0\)

=>\(\left(2\cdot\sin\left(\frac{x}{2}\right)-1\right)\left(\sin\left(\frac{x}{2}\right)-1\right)=0\)

TH1: \(\sin\left(\frac{x}{2}\right)-1=0\)

=>\(\sin\left(\frac{x}{2}\right)=1\)

=>\(\frac{x}{2}=\frac{\pi}{2}+k2\pi\)

=>\(x=\pi+k4\pi\)

TH2: \(2\cdot\sin\left(\frac{x}{2}\right)-1=0\)

=>\(\sin\left(\frac{x}{2}\right)=\frac12\)

=>\(\left[\begin{array}{l}\frac{x}{2}=\frac{\pi}{6}+k2\pi\\ \frac{x}{2}=\pi-\frac{\pi}{6}+k2\pi=\frac56\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{3}+k4\pi\\ x=\frac53\pi+k4\pi\end{array}\right.\)

\(\int\limits^{\dfrac{\pi}{4}}_{\dfrac{\pi}{8}}\dfrac{dx}{sin^2x.cos^2x}=\int\limits^{\dfrac{\pi}{4}}_{\dfrac{\pi}{8}}\dfrac{2d\left(2x\right)}{sin^22x}=-2cot2x|^{\dfrac{\pi}{4}}_{\dfrac{\pi}{8}}=...\) 

\(\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\dfrac{cos2xdx}{sin^2x.cos^2x}=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\dfrac{cos^2x-sin^2x}{sin^2x.cos^2x}dx=\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\left(\dfrac{1}{sin^2x}-\dfrac{1}{cos^2x}\right)dx=\left(-cotx-tanx\right)|^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\)

\(\int\limits^{\dfrac{\pi}{3}}_0\dfrac{cos3x}{cosx}dx=\int\limits^{\dfrac{\pi}{3}}_0\dfrac{4cos^3x-3cosx}{cosx}dx=\int\limits^{\dfrac{\pi}{3}}_0\left(4cos^2x-3\right)dx\)

\(=\int\limits^{\dfrac{\pi}{3}}_0\left(2cos2x-1\right)dx=\left(sin2x-x\right)|^{\dfrac{\pi}{3}}_0=...\)

Đặt \(t=\pi-x\Rightarrow dx=-dt\)

\(I=\int\limits^0_{\pi}\dfrac{\left(\pi-t\right)sint}{sin^2t+3}.-dt=\int\limits^{\pi}_0\dfrac{\left(\pi-t\right)sint}{sin^2t+3}dt=\int\limits^{\pi}_0\dfrac{\left(\pi-x\right)sinx}{sin^2x+3}dx\)

\(\Rightarrow2I=I+I=\int\limits^{\pi}_0\left(\dfrac{xsinx}{sin^2x+3}+\dfrac{\left(\pi-x\right)sinx}{sin^2x+3}\right)dx=\pi\int\limits^{\pi}_0\dfrac{sinx}{sin^2x+3}dx\)

\(=-\pi\int\limits^{\pi}_0\dfrac{d\left(cosx\right)}{4-cos^2x}=-\dfrac{\pi}{4}ln\left|\dfrac{2+cosx}{2-cosx}\right||^{\pi}_0=\dfrac{\pi.ln3}{2}\)

\(\Rightarrow I=\dfrac{\pi.ln3}{4}\)

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:

\(\int \frac{\tan ^2x-\cos ^2x}{\sin ^2x}dx=\tan x+c-(-\cot x-x+c)=\tan x+\cot x+x+c\)

\(\Rightarrow \int ^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\tan ^2x-\cos ^2x}{\sin ^2x}dx=\frac{4\sqrt{3}}{3}+\frac{\pi}{3}-\frac{4\sqrt{3}}{3}-\frac{\pi}{6}=\frac{\pi}{6}\)

a: \(5-2\cdot cos^2x\cdot\sin^2x\)

\(=5-2\cdot\left(\sin x\cdot cosx\right)^2\)

\(=5-2\cdot\left\lbrack\frac12\cdot2\cdot\sin x\cdot cosx\right\rbrack^2=5-2\cdot\left\lbrack\frac12\cdot\sin2x\right\rbrack^2\)

\(=5-2\cdot\frac14\cdot\sin^22x=-\frac12\cdot\sin^22x+5\)

\(0\le\sin^22x\le1\)

=>\(0\ge-\frac12\sin^22x\ge-\frac12\)

=>\(0+5\ge-\frac12\sin^22x+5\ge-\frac12+5\)

=>\(5\ge-\frac12\sin^22x+5\ge\frac92\)

=>\(\frac92\le-\frac12\sin^22x+5\le5\)

=>\(\sqrt{\frac92}\le\sqrt{-\frac12\cdot\sin^22x+5}\le\sqrt5\)

=>\(\frac{3\sqrt2}{2}\le\sqrt{-\frac12\cdot\sin^22x+5}\le\sqrt5\)

=>\(\frac{2}{3\sqrt2}\ge\frac{1}{\sqrt{-\frac12\cdot\sin^22x+5}}\ge\frac{1}{\sqrt5}\)

=>\(\frac{2\cdot4}{3\sqrt2}\ge\frac{1\cdot4}{\sqrt{-\frac12\cdot\sin^22x+5}}\ge\frac{1\cdot4}{\sqrt5}\)

=>\(\frac{4\sqrt2}{3}\ge y\ge\frac{4}{\sqrt5}\)

=>\(y_{\max}=\frac{4\sqrt2}{3}\) khi \(-\frac12\cdot\sin^22x+5=\frac92\)

=>\(-\frac12\cdot\sin^22x=-\frac12\)

=>\(\sin^22x=1\)

=>\(cos^22x=0\)

=>cos2x=0

=>\(2x=\frac{\pi}{2}+k\pi\)

=>\(x=\frac{\pi}{4}+\frac{k\pi}{2}\)

\(y_{\min}=\frac{4}{\sqrt5}\) khi \(-\frac12\cdot\sin^22x+5=5\)

=>\(\sin^22x=0\)

=>sin 2x=0

=>\(2x=k\pi\)

=>\(x=\frac{k\pi}{2}\)

b: \(f\left(x\right)=3\cdot\sin^2x+5\cdot cos^2x-4\cdot cos2x-2\)

\(=3\left(1-cos^2x\right)+5\cdot cos^2x-4\left(2\cdot cos^2x-1\right)-2\)

\(=3-3\cdot cos^2x+5\cdot cos^2x-8\cdot cos^2x+4-2=-6\cdot cos^2x+5\)

Ta có: \(0<=cos^2x\le1\)

=>\(0\ge-6\cdot cos^2x\ge-6\)

=>\(0+5\ge-6\cdot cos^2x+5\ge-6+5\)

=>5>=y>=-1

Do đó: \(y_{\min}=-1\) khi \(-6\cdot cos^2x+5=-1\)

=>\(-6\cdot cos^2x=-6\)

=>\(cos^2x=1\)

=>\(\sin^2x=0\)

=>sin x=0

=>\(x=k\pi\)

y max=5 khi \(-6\cdot cos^2x+5=5\)

=>\(-6\cdot cos^2x=0\)

=>cosx=0

=>\(x=\frac{\pi}{2}+k\pi\)