Question

2cos^2(x/8) найти интеграл

Answer 1

$\LARGE \\ \int\begin{pmatrix} 2\cos^2{x\over8} \end{pmatrix}\mathrm{dx}=2\int{\cos^2{x\over8}}\mathrm{dx}=\begin{vmatrix} {x\over8}=t\\ \mathrm{dx}=8\mathrm{dt} \end{vmatrix}=16\int\cos^2{t}\mathrm{dt}=16\int\begin{pmatrix} {1+\cos{2t}\over2} \end{pmatrix}\mathrm{dt}=8\int\mathrm{dt}+8\int\cos{2t}\mathrm{dt}=8t+4\int\cos{2t}\mathrm{d(2t)}=8t+4\sin{2t}+C=x+4\sin{x\over4}+C$

Answer 2

$\displaystyle \int2cos^2(\frac{x}{8})dx=2\int cos^2(\frac{x}{8})dx=\int(1+cos(\frac{x}{4}))dx=\int dx+\\+4\int cos(\frac{x}{4})d(\frac{x}{4})=x+4sin\frac{x}{4}+C$