Cảm ơn Hương Trà

\(I=\int\limits^{\dfrac{\pi}{2}}_0\left(1+cosx+x.cosx\right)e^{sinx}dx=\int\limits^{\dfrac{\pi}{2}}_0e^{sinx}dx+\int\limits^{\dfrac{\pi}{2}}_0\left(x+1\right).cosx.e^{sinx}dx=I_1+I_2\)

Xét \(I_2\), đặt \(\left\{{}\begin{matrix}u=x+1\\dv=cosx.e^{sinx}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=e^{sinx}\end{matrix}\right.\)

\(\Rightarrow I_2=\left(x+1\right).e^{sinx}|^{\dfrac{\pi}{2}}_0-\int\limits^{\dfrac{\pi}{2}}_0e^{sinx}dx=\left(\dfrac{\pi}{2}+1\right)e-1-I_1\)

\(\Rightarrow I=I_1+\left(\dfrac{\pi}{2}+1\right)e-1-I_1=\left(\dfrac{\pi}{2}+1\right)e-1\)

Bạn xem lại xem có type thiếu đề không? \((x+\frac{\pi}{6})\) có sin hay cos, tan ở phía trước không?

Sin nha

\(\int\limits^a_b\left(sinx+cosx\right)dx=\left(sinx-cosx\right)|^a_b=sina-cosa-sinb+cosb=m\)

\(\int\limits^b_a\left(sinx-cosx\right)dx=\left(-cosx-sinx\right)|^b_a=-cosa-sina+cosb+sinb=n\)

\(\Rightarrow\left\{{}\begin{matrix}m+n=-2\left(cosa-cosb\right)\\m-n=2\left(sina-sinb\right)\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}cosa-cosb=-\dfrac{m+n}{2}\\sina-sinb=\dfrac{m-n}{2}\end{matrix}\right.\)

\(I=\int\limits^b_asin\left(x+\dfrac{\pi}{6}\right)dx=-cos\left(x+\dfrac{\pi}{6}\right)|^b_a=cos\left(a+\dfrac{\pi}{6}\right)-cos\left(b+\dfrac{\pi}{6}\right)\)

\(=cosa.cos\left(\dfrac{\pi}{6}\right)-sina.sin\left(\dfrac{\pi}{6}\right)-cosb.cos\left(\dfrac{\pi}{6}\right)+sinb.sin\left(\dfrac{\pi}{6}\right)\)

\(=\dfrac{\sqrt{3}}{2}\left(cosa-cosb\right)-\dfrac{1}{2}\left(sina-sinb\right)\)

\(=\dfrac{-\sqrt{3}}{4}\left(m+n\right)-\dfrac{1}{4}\left(m-n\right)\)

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