Ta có \(I=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{\ln2.\ln\left(2\tan x\right)}{\sin2x.\ln\left(2\tan x\right)}dx=\ln2\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x.\ln\left(2\tan x\right)}+\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x}\)

Tính \(\ln2\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x.\ln\left(2\tan x\right)}=\frac{\ln2}{2}\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{d\left[\ln\left(2\tan x\right)\right]}{\ln2\left(2\tan x\right)}=\frac{\ln2}{2}\left[\ln\left(\ln\left(2\tan x\right)\right)\right]|^{\frac{\pi}{3}}_{\frac{\pi}{4}}=\frac{\ln2}{2}.\ln\left(\frac{\ln2\sqrt{3}}{\ln2}\right)\)

Tính \(\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x}=\frac{1}{2}\ln\left(\tan x\right)|^{\frac{\pi}{3}}_{\frac{\pi}{4}}=\frac{1}{2}\ln\sqrt{3}\)

Vậy \(I=\frac{\ln2}{2}\ln\left(\frac{\ln2\sqrt{3}}{\ln2}\right)+\frac{1}{2}\ln\sqrt{3}\)

\(I=\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{\cos2x+3\cos x+2}dx=\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{2\cos^2x+3\cos x+1}dx\)

Đặt \(\cos x=t\Rightarrow dt=-\sin dx\)

Với \(x=0\Rightarrow t=1\)

Với \(x=\frac{\pi}{2}\Rightarrow t=0\)

\(I=\int\limits^1_0\frac{dt}{2t^2+3t+1}=\int\limits^1_0\frac{dt}{\left(2t+1\right)\left(t+1\right)}=2\int\limits^1_0\left(\frac{1}{2t+1}+\frac{1}{2t+1}\right)dt\)

  \(=\left(\ln\frac{2t+1}{2t+1}\right)|^1_0=\ln\frac{3}{2}\)

\(\int\limits^{\frac{\pi}{4}}_0\frac{x}{\cos^2}dx=\int\limits^{\frac{\pi}{4}}_0x.d\left(\tan x\right)=x.\tan|^{\frac{\pi}{4}}_0-\int\limits^{\frac{\pi}{4}}_0\tan xdx=\frac{\pi}{4}+\ln\left(\cos x\right)|^{\frac{\pi}{4}}=\frac{\pi}{4}-\frac{1}{2}\ln2\)

\(=\frac{1}{2}\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\ln\left(\tan x\right)d\left[\ln\left(\tan x\right)\right]=\frac{1}{4}\left[\ln^2\left(\tan x\right)\right]|^{\frac{\pi}{3}}_{\frac{\pi}{4}}=\frac{1}{4}\left(\ln^2\sqrt{3}-0\right)=\frac{1}{16}\ln^23\)

Đặt \(t=\tan x\Rightarrow\begin{cases}dt=\frac{dt}{\cos^2}=\left(1+t^2\right)dx\rightarrow dx=\frac{dt}{1+t^2}\\x=\frac{\pi}{4}\rightarrow t=1;x=\frac{\pi}{3}\rightarrow t=\sqrt{3}\end{cases}\)

Khi đó : \(I=\int\limits^{\sqrt{3}}_1\frac{\ln t}{\frac{2t}{1+t^2}}.\frac{dt}{1+t^2}=\frac{1}{2}\int\limits^{\sqrt{3}}_1\frac{\ln t}{t}dt=\frac{1}{2}J\left(1\right)\)

\(J=\int\limits^{\sqrt{3}}_1\frac{\ln t}{t}dt=\int\limits^{\sqrt{3}}_1\ln.d\left(\ln t\right)=\frac{1}{2}\ln^2t|^{\sqrt{3}}_1=\frac{1}{2}\left(\ln^2\sqrt{3}-0\right)=\frac{1}{8}\ln^23\)

Thay vào (1) ta có : \(I=\frac{1}{16}\ln^23\)

Mình giải giúp b câu 1 này

Ở phần mẫu bạn biến đổi \(cos^2xsin^2x=\frac{1}{4}\left(4cos^2xsin^2x\right)=\frac{1}{4}sin^22x\)

Đặt t = sin2x => \(d\left(t\right)=2cos2xdx\)

Đổi cận \(x=\frac{\pi}{4}=>t=1\) \(x=\frac{\pi}{3}=>t=\frac{\sqrt{3}}{2}\)

Ta có biểu thức trên sau khi đổi biến và cận

\(\int\limits^{\frac{\sqrt{3}}{2}}_1\frac{\frac{1}{2}dt}{\frac{1}{4}t^2}=\int\limits^{\frac{\sqrt{3}}{2}}_1\frac{2}{t^2}dt=\left(-\frac{2}{t}\right)\)lấy cận từ 1 đến \(\frac{\sqrt{3}}{2}\) \(=-\frac{2}{\frac{\sqrt{3}}{2}}-\left(-\frac{2}{1}\right)=2-4\frac{\sqrt{3}}{3}\) => a=2 và b=-4/3 vậy A=2/3 nhé

Câu 1)

Ta có:

\(I=\int ^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{\cos 2x}{\cos^2 x\sin^2 x}dx=\int ^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{\cos^2x-\sin ^2x}{\cos^2 x\sin^2 x}dx\)

\(=\int ^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin^2 x}-\int ^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\cos ^2x}=-\int ^{\frac{\pi}{3}}_{\frac{\pi}{4}}d(\cot x)-\int ^{\frac{\pi}{3}}_{\frac{\pi}{4}}d(\tan x)\)

\(=-\left ( \frac{\sqrt{3}}{3}-1 \right )-(\sqrt{3}-1)=2-\frac{4}{3}\sqrt{3}\Rightarrow a+b=\frac{2}{3}\)

Câu 2)

\(I=\underbrace{\int ^{\frac{\pi}{2}}_{0}\sin ^2xdx}_{A}+\underbrace{\int ^{\frac{\pi}{2}}_{0}\frac{\sin x\cos 2xdx}{\sqrt{1+3\cos x}}}_{B}\)

Có \(A=\int ^{\frac{\pi}{2}}_{0}\frac{1-\cos 2x}{2}dx=\)\(\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\left ( \frac{x}{2}-\frac{\sin 2x}{4} \right )=\frac{\pi}{4}\)

\(B=-\int ^{\frac{\pi}{2}}_{0}\frac{(2\cos ^2x-1)d(\cos x)}{\sqrt{1+3\cos x}}\). Ta đặt \(\sqrt{1+3\cos x}=t\)

\(B=B=\int ^{2}_{1}\frac{\left [ \frac{2(t^2-1)^2}{9}-1\right ]d\left ( \frac{t^2-1}{3} \right )}{t}=\frac{2}{27}\int ^{2}_{1}\left ( 2t^4-4t^2-7 \right )dt\)

\(=\left.\begin{matrix} 2\\ 1\end{matrix}\right|\frac{2}{27}\left ( \frac{2t^5}{5}-\frac{4t^3}{3}-7t \right )=\frac{-118}{405}\)

\(\left\{\begin{matrix} a=\frac{1}{4}\\ b=-118\\ c=405\end{matrix}\right.\Rightarrow a+b+c=287,25\)

Bài này mà ngồi trong phòng thi mà giải tay thì chết cmnr. Bạn lên youtube xem anh theluc giải bằng casio cho nhanh.