Đặt \(u=x^2\rightarrow du=2xdx,dv=\cos xdx\rightarrow v=\sin x\)

Do đó : 

\(I=x^2.\sin x|^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_02x.\sin xdx=\frac{\pi^2}{4}+\int\limits^{\frac{\pi}{2}}_0x.d\left(\cos x\right)=\frac{\pi^2}{4}+\left(x.\cos x|^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_0\cos x\right)\)

\(=\frac{\pi^2}{4}+\left(0-\sin|^{\frac{\pi}{2}}_0\right)=\frac{\pi^2-4}{4}\)

\(I=\int\limits^{\pi}_0\left(x^2-x\sin x\right)dx=\frac{x^3}{3}|^{\pi}_0-\int^{\pi}_0x\sin xdx=\frac{\pi^3}{3}-\int\limits^{\pi}_0x\sin xdx\)

Tính \(I_1=\int\limits^{\pi}_0x\sin xdx\)

Đặt \(\begin{cases}u=x\\dv=\sin xdx\end{cases}\)\(\Rightarrow\begin{cases}du=dx\\v=-\cos x\end{cases}\)

\(\Rightarrow I_1=-x\cos x|^{\pi}_0+\int\limits^{\pi}_0\cos xdx=\pi+\sin x|^{\pi}_0=\pi\)

\(\Rightarrow I=\frac{\pi^3}{3}-\pi\)

\(I=\int\limits^{\frac{\pi}{2}}_0\left(2x-1\right)\cos^2xdx=\int\limits^{\frac{\pi}{2}}_0\left(2x-1\right)\left(\frac{1+\cos2x}{2}\right)dx=\int\limits^{\frac{\pi}{2}}_0\left(x-\frac{1}{2}\right)dx+\frac{1}{2}\int\limits^{\frac{\pi}{2}}_0\left(2x-1\right)\cos2xdx\)

 \(=\left(\frac{1}{2}x^2-\frac{1}{2}x\right)|^{\frac{\pi}{2}}_0+\frac{1}{2}\int\limits^{\frac{\pi}{2}}_0\left(2x-1\right)d\left(\sin2x\right)=\frac{\pi^2}{8}-\frac{\pi}{4}+\frac{1}{2}\left[\left(2x-1\right)\sin2x|^{\frac{\pi}{2}}_0-\int\limits^{^{\frac{\pi}{2}}_0}_0\sin2x.2dx\right]\)

 \(=\frac{\pi^2}{8}-\frac{\pi}{4}+\left(0+\cos2x|^{\frac{\pi}{2}}_0\right)=\frac{\pi^2}{8}-\frac{\pi}{4}-1\)

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

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

d)

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

Ta có:

\(\int ^{\frac{\pi}{4}}_{0}dx=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|x=\frac{\pi}{4}\)

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

\(=\ln|\frac{\pi\sqrt{2}}{8}+\frac{\sqrt{2}}{2}|\)

Suy ra \(D=\frac{\pi}{4}+\ln|\frac{\pi\sqrt{2}}{8}+\frac{\sqrt{2}}{2}|\)

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

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

Với \(x=0\Rightarrow t=1\)

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

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

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

Cách này hơi dài chút, nhưng nếu nghĩ ra cách hay hơn mình sẽ đề xuất nhe!

\(=\int\sin^5x.\left(2\sin x\cos x\right)^3.2xdx=16\int x.\sin^8x\cos^3xdx\)

\(\left\{{}\begin{matrix}u=x\\dv=\sin^8x.\cos^3xdx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=dx\\v=\int\sin^8x.\cos^3xdx\end{matrix}\right.\)

\(I_1=\int\sin^8x\cos^3xdx=\int\sin^8x.\cos^2x.\cos xdx=\int\sin^8x.\left(1-\sin^2x\right)\cos xdx\)

\(t=\sin x\Rightarrow dt=\cos xdx\Rightarrow\int\sin^8x\left(1-\sin^2x\right)\cos xdx=\int(t^8-t^{10})dt=\dfrac{1}{9}t^9-\dfrac{1}{11}t^{11}=\dfrac{1}{9}\sin^9x-\dfrac{1}{11}\sin^{11}x\)

\(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=\dfrac{1}{9}\sin^9x-\dfrac{1}{11}\sin^{11}x\end{matrix}\right.\)

\(\Rightarrow\dfrac{I}{16}=x.\left(\dfrac{1}{9}\sin^9x-11\sin^{11}x\right)-\int\left(\dfrac{1}{9}\sin^9x-\dfrac{1}{11}\sin^{11}x\right)dx\)

\(I_2=\int\left(\dfrac{1}{9}\sin^9x-\dfrac{1}{11}\sin^{11}x\right)dx=\dfrac{1}{9}\int\sin^9xdx-\dfrac{1}{11}\int\sin^{11}xdx\)

À thế này là xong rồi còn gì :) Bạn tự làm nốt nhé