ạ
\(I_7=\int\limits^3_0x^2\sqrt{10-x^2}xdx\)
Đặt \(\sqrt{10-x^2}=t\Rightarrow x^2=10-t^2\Rightarrow xdx=tdt\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=\sqrt{10}\\x=3\Rightarrow t=1\end{matrix}\right.\)
\(I_7=\int\limits^1_{\sqrt{10}}\left(10-t^2\right)t.tdt=\int\limits^{\sqrt{10}}_1\left(t^4-10t^2\right)dt\)
\(=\left(\dfrac{1}{5}t^5-\dfrac{10}{3}t^3\right)|^{\sqrt{10}}_1\) (tới đây bạn tự tính ra kết quả nhé)
\(I_8=\int\limits^{\dfrac{\pi}{4}}_02sinx.cosx.cosxdx=-2\int\limits^{\dfrac{\pi}{4}}_0cos^2x.d\left(cosx\right)\)
\(=-\dfrac{2}{3}cos^3x|^{\dfrac{\pi}{4}}_0=...\)
\(I_9=\int\limits^{\dfrac{\pi}{2}}_0\dfrac{sinx}{1+3cosx}dx\)
Đặt \(u=cosx\Rightarrow du=-sinx.dx\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow u=1\\x=\dfrac{\pi}{2}\Rightarrow u=0\end{matrix}\right.\)
\(I_9=\int\limits^0_1\dfrac{-du}{1+3u}=\dfrac{1}{3}\int\limits^1_0\dfrac{d\left(3u+1\right)}{3u+1}=\dfrac{1}{3}ln\left(3u+1\right)|^1_0=\dfrac{1}{3}ln10\)
\(I_{10}=\int\limits^{\dfrac{\pi}{2}}_0sin^4x.\left(1-sin^2x\right)^2cosxdx\)
Đặt \(u=sinx\Rightarrow du=cosxdx\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow u=0\\x=\dfrac{\pi}{2}\Rightarrow u=1\end{matrix}\right.\)
\(I_{10}=\int\limits^1_0u^4\left(1-u^2\right)du=\int\limits^1_0\left(u^8-2u^6+u^4\right)du\)
\(=\left(\dfrac{1}{9}u^9-\dfrac{2}{7}u^7+\dfrac{1}{5}u^5\right)|^1_0=...\)
\(I_{11}=\int\limits^{\dfrac{\pi}{2}}_0\left(2x-1\right)cosxdx\)
Đặt \(\left\{{}\begin{matrix}u=2x-1\\dv=cosxdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2dx\\v=sinx\end{matrix}\right.\)
\(\Rightarrow I_{11}=\left(2x-1\right)sinx|^{\dfrac{\pi}{2}}_0-2\int\limits^{\dfrac{\pi}{2}}_0sinxdx\)
\(=\pi-1+2cosx|^{\dfrac{\pi}{2}}_0=\pi-3\)
\(I_{12}=\int\limits^1_0\dfrac{e^xdx}{1+e^x}=\int\limits^1_0\dfrac{d\left(e^x+1\right)}{e^x+1}\)
\(=ln\left(e^x+1\right)|^1_0=ln\left(e+1\right)-ln2=ln\left(\dfrac{e+1}{2}\right)\)
\(I_{13}=\int\limits^2_0xe^{2x}dx\)
Đặt \(\left\{{}\begin{matrix}u=x\\dv=e^{2x}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=\dfrac{1}{2}e^{2x}\end{matrix}\right.\)
\(I_{13}=\dfrac{1}{2}xe^{2x}|^2_0-\dfrac{1}{2}\int\limits^2_0e^{2x}dx=\left(\dfrac{1}{2}xe^{2x}-\dfrac{1}{4}e^{2x}\right)|^2_0=...\)
\(I_{14}=\int\limits^e_1\dfrac{\sqrt{1+ln^2x}}{x}dx\)
Đặt \(lnx=u\Rightarrow du=\dfrac{dx}{x}\) ; \(\left\{{}\begin{matrix}x=1\Rightarrow u=0\\x=e\Rightarrow u=1\end{matrix}\right.\)
\(I_{14}=\int\limits^1_0\sqrt{1+u^2}du\)
Đặt \(\sqrt{1+u^2}=t+u\Leftrightarrow1+u^2=u^2+t^2+2ut\)
\(\Rightarrow1-t^2=2ut\Rightarrow u=\dfrac{1-t^2}{2t}\Rightarrow du=-\dfrac{t^2+1}{2t^2}dt\) ; \(\left\{{}\begin{matrix}u=0\Rightarrow t=1\\u=1\Rightarrow t=\sqrt{2}-1\end{matrix}\right.\)
\(I_{14}=-\int\limits^{\sqrt{2}-1}_1\left(t+\dfrac{1-t^2}{2t}\right)\left(\dfrac{t^2+1}{2t^2}\right)dt\)
\(=\dfrac{1}{4}\int\limits^1_{\sqrt{2}-1}\left(t+\dfrac{2}{t}+\dfrac{1}{t^3}\right)dt=\dfrac{1}{4}\left(\dfrac{t^2}{2}+2lnt-\dfrac{1}{2t^2}\right)|^1_{\sqrt{2}-1}=...\)
giúp mình với ậ