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

a/  \(W=\dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2=\dfrac{1}{2}k\Delta l^2\)

\(\Leftrightarrow kx^2+mv^2=k\Delta l^2\Leftrightarrow v=\sqrt{\dfrac{k\Delta l^2-kx^2}{m}}=\sqrt{\dfrac{40.0,02^2-40x^2}{0,4}}\left(m/s\right)\)

b/ \(v_{max}\Leftrightarrow\dfrac{40.0,02^2-40x^2}{0,4}\left(max\right)\Leftrightarrow x=0\) => khi nó ở VTCB

\(\Rightarrow v_{max}=\dfrac{40.0,02^2}{0,4}\left(m/s\right)\)

Ta có: \(\overrightarrow{P}=m.\overrightarrow{a}\Rightarrow Ox:mg\sin\alpha=m.a\Leftrightarrow a=g\sin\alpha\left(m/s^2\right)\)

\(v^2-v_0^2=2aS\Leftrightarrow S=-\dfrac{v_0^2}{2a}=-\dfrac{v_0^2}{2.g\sin\alpha}=...\left(m\right)\)

\(\left\{{}\begin{matrix}\left(\sqrt{5}\right)^{2x}-\left(\sqrt{5}\right)^{3y}=\left(3y-2x\right)\left(6xy+12\right)\left(1\right)\\4x^2+9y^2=16\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Rightarrow4x^2+9y^2-4=12\) the vo (1)

\(\Rightarrow\left(\sqrt{5}\right)^{2x}-\left(\sqrt{5}\right)^{3y}=\left(3y-2x\right)\left(6xy+4x^2+9y^2-4\right)\)

\(\Leftrightarrow\left(\sqrt{5}\right)^{2x}-\left(\sqrt{5}\right)^{3y}=27y^3-8x^3-12y+8x\)

\(\Leftrightarrow\left(\sqrt{5}\right)^{2x}+\left(2x\right)^3-4.\left(2x\right)=\left(\sqrt{5}\right)^{3y}+\left(3y\right)^3-4.\left(3y\right)\left(3\right)\)

Xét hàm số \(f\left(t\right)=\left(\sqrt{5}\right)^{2t}+\left(2t\right)^3-4.2t\)  đồng biến trên R

\(\Rightarrow\left(3\right):f\left(2x\right)=f\left(3y\right)\Leftrightarrow\left\{{}\begin{matrix}2x=3y\\4x^2+9y^2=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{2}\\y=\dfrac{2\sqrt{2}}{3}\end{matrix}\right.\)

Các bạn rất hay mắc một lỗi rất cơ bản, đó chính là đặt x ra ngoài căn nhưng ko phải là |x|? 

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{\left|x\right|\sqrt{\dfrac{4x^2}{x^2}+\dfrac{2x}{x^2}-\dfrac{1}{x^2}}-x}{3x-2}=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{4}-1}{3}=-1\)

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

\(\Rightarrow\int xln\left(x+1\right)dx=\dfrac{1}{2}x^2.ln\left(x+1\right)-\dfrac{1}{2}\int\dfrac{x^2}{x+1}dx\)

\(\int\dfrac{x^2dx}{x+1}=\int\left(x-1\right)dx+\int\dfrac{dx}{x+1}\)

P/s: Tất cả đã về dạng cơ bản, bạn tự làm nốt ạ

\(\int\dfrac{lnx}{x\left(2ln^2x-1\right)^3}dx\)

\(t=2ln^2x-1\Rightarrow dt=\dfrac{4}{x}lnxdx\Rightarrow dx=\dfrac{x.dt}{4lnx}\)

\(\Rightarrow\int\dfrac{lnx}{x\left(2ln^2x-1\right)^3}dx=\int\dfrac{lnx}{x\left(2ln^2x-1\right)^3}.\dfrac{xdt}{4lnx}=\dfrac{1}{4}\int\dfrac{dt}{t^3}=\dfrac{1}{4}.\left(-\dfrac{1}{2}\right).t^{-2}=-\dfrac{1}{8\sqrt{2ln^2x-1}}\)

\(\int\dfrac{dx}{x^3+x}=\int\dfrac{dx}{x\left(x^2+1\right)}\)

\(t=x^2+1\Rightarrow dt=2xdx\Rightarrow\int\dfrac{dx}{x\left(x^2+1\right)}=\int\dfrac{dt}{2x^2t}=\dfrac{1}{2}\int\dfrac{dt}{\left(t-1\right).t}\)

\(\dfrac{1}{\left(t-1\right).t}=\dfrac{1}{t-1}-\dfrac{1}{t}\)

\(\Rightarrow\int\dfrac{dt}{\left(t-1\right)t}=\int\left(\dfrac{1}{t-1}-\dfrac{1}{t}\right)dt=\int\dfrac{dt}{t-1}-\int\dfrac{dt}{t}=ln\left|t-1\right|-ln\left|t\right|=ln\left|x^2\right|-ln\left|x^2+1\right|\)