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Объяснение:

4.

[tex]a)\ \int\limits {(1-tg(3x))^2} \, dx=\int\limits {(tg^2(3x)-2*tg(3x)+1)} \, dx=\left | {{u=3x\ \ \ \ x=\frac{u}{3} } \atop {dx=\frac{du}{3} }} \right. |=\\ =\frac{1}{3}*( \int\limits {tg^2(u)} \, du-2*\int\limits {tg(u)} \, du+\int du).\\1)\ \int tg^2(u)du=\int \frac{sin^2(u)}{cos^2(u)} du=\int\frac{sin^2(u)+cos^2(u)-cos^2(u)}{cos^2(u)}du=\int \frac{1-cos^2(u)}{cos^2(u)}du=\\ =\int (\frac{1}{cos^2(u)} -1)du=\int\frac{du}{cos^2(u)} -\int du =tg(u)-u=tg(3x)-3x.\\[/tex]

[tex]2)\ \int tg(u)du=\int \frac{sin(u)}{cos(u)} du=\left | {{t=cos(u)} \atop {dt=-sinudu}} \right.|= -\int \frac{dt}{t} =-ln|t|+C=\\=-ln|cos(u)|+C=-ln|cos(3x)|+C.[/tex]

[tex]3)\ \int du=u+C=3x+C.\ \ \ \ \Rightarrow[/tex]

[tex]\frac{1}{3} *(tg(3x)-3x-2*(-ln|cos(3x)|)+3x)+C=\frac{2*ln|cos(3x)|}{3}+\frac{tg(3x)}{3} +C.[/tex]

[tex]b)\ \int sin(2x+1)*sin(3x-5)dx=\int \frac{2*sin(2x+1)*sin(3x-5)}{2}=\\=\frac{1}{2}*\int (cos( 3x-5-2x-1)-sin(3x+5+2x+1))dx=\\=\frac{1}{2}*\int(cos(x-6)-cos(5x-4) )dx=\frac{1}{2}*(\int cos(x-6)-\int cos(5x-4)).\\[/tex]

[tex]1)\ \int cos(x-6)dx=\left | {{u=x-6} \atop {du=dx}} \right.|=\int cos(u)du =sin(u) +C=sin(x-6)+C.\\2)\ \int cos(5x-4)dx=\left \{ {{u=5x-4\ \ \ \ \ du=5dx} \atop {dx=\frac{du}{5} }} \right. |=\frac{1}{5}* \int cos(u)du=\\=\frac{sin(u)}{5} +C=\frac{sin(5x-4)}{5}+C.\ \ \ \ \ \Rightarrow\\[/tex]

[tex]\frac{1}{2}*(sin(x-6)-\frac{sin(5x-4)}{5} )+C=\frac{sin(x-6)}{2}-\frac{sin(5x-4)}{10} +C.[/tex]

11.

[tex]a)\ \int \frac{xdx}{y-1+\sqrt{x-1} }=\left | {{u=x-1\ \ \ \ \ x=u+1} \atop {dx=du}} \right.|=\int \frac{u+1}{1+\sqrt{u} } du=\left | {{v=\sqrt{u}\ \ \ \ v^2=u } \atop {du=2vdv}} \right.| =\int \frac{(v^2+1)*2v}{v+1}dv=\\ = \left | {{t=v+1\ \ \ \ v=t-1} \atop {dv=dt}} \right. |=2*\int\frac{((t-1)^2+1)*(t-1)}{t}dt=2*\int \frac{(t^2-2t+1+1)*(t-1)}{t} dt=\\[/tex]

[tex]=2*\int \frac{t^3-2t^2+2t-t^2+2t-1}{t}dt=2*\int\frac{t^3-3t^2+4t-2}{t} =2*\int(t^2-3t+4-\frac{2}{t} )dt=\\ =2*(\frac{t^3}{3}-\frac{3t^2}{2}+4t-2*ln|t|)+C=\frac{2t^3}{3}-3t^2+8t-4ln|t|+C=\\ =\frac{2*(v+1)^3}{3}-3*(v+1)^2+8*(v+1)-4*ln|v+1|+C=\\ =\frac{2*(v+1)^3}{3}-3v^2-6v-3+8v+8-4*ln|v+1|+C=\\[/tex]

[tex]=\frac{2*(v+1)^3 }{3 } -3v^2+2v+5-4*ln|v+1|+C=\\=\frac{2*(\sqrt{u}+1)^3 }{3} -3*(\sqrt{u})^2+2*\sqrt{u} +5-4*ln|\sqrt{u}+1|+C= \\ =\frac{2*(\sqrt{u}+1)^3 }{3} -3*u+2*\sqrt{u} +5-4*ln|\sqrt{u}+1|+C= \\ =\frac{2(\sqrt{x-1}+1 )^3}{3} -3(x-1) +2\sqrt{x-1} +5-4ln|\sqrt{x-1}+1|+C=\\ =-3x+\frac{2(\sqrt{x-1}+1)^3 }{3} +2\sqrt{x-1}+8-4ln|\sqrt{x-1}+1|+C.[/tex]