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Ответ:

16)  Применим формулу произведения косинусов .

[tex]\bf \displaystyle \int cos^2ax\cdot cos^2bx\, dx=\int (cosax\cdot cosbx)^2\, dx=\\\\\\=\int \frac{1}{4}\Big(cos(ax+bx)+coa(ax-bx)\Big)^2\, dx=\\\\\\=\frac{1}{4}\int \Big(cos^2(a+b)x+cos^2(a-b)x-2cos(a+b)x\cdot cos(a-b)x\Big)\, dx=\\\\\\=\frac{1}{4}\int \frac{1+cos\frac{(a+b)}{2}x}{2}\, dx+\frac{1}{4}\int \frac{1+cos\frac{a-b}{2}x}{2}\, dx-\frac{1}{4}\int \Big(cos(2ax)+cos(2bx)\Big)\, dx=[/tex]

[tex]\bf \displaystyle =\frac{1}{8}\Big(x+\frac{2}{a+b}sin\frac{(a+b)x}{2}\Big)+\frac{1}{8}\Big(x+\frac{2}{a-b}sin\frac{(a-b)x}{2}\Big)-\frac{1}{8a}sin(2ax)-\\\\\\-\frac{1}{8a}sin(2bx)+C[/tex]  

18)  Замена переменной .

[tex]\bf \displaystyle \int \frac{\sqrt[3]{\bf 2x+1}}{1+\sqrt[3]{\bf 2x+1}}\, dx=\Big[\ t=\sqrt[3]{\bf 2x+1}\ ,\ t^3=2x+1\ ,\ x=\frac{t^3-1}{2}\ ,\\\\\\dx=\frac{3t^2}{2}\, dt\ \Big]=\int \frac{t\cdot 3t^2}{2\, (1+t)}\, dt=\frac{3}{2}\int \frac{t^3}{t+1}\, dt=\\\\\\=\frac{3}{2}\int \Big(t^2-t+1-\frac{1}{t+1}\Big)\, dt=\frac{3}{2}\Big(\frac{t^3}{3}-\frac{t^2}{2}+t-ln|\, t+1\, |\Big)+C=[/tex]

[tex]\bf \displaystyle =\frac{2x+1}{2}-\frac{3\sqrt[3]{\bf (2x+1)^2}}{4} +\frac{3\sqrt[3]{\bf 2x+1}}{2}-\frac{3}{2}\, ln\Big|\, \sqrt[3]{\bf 2x+1}+1\, \Big|+C[/tex]