Помогите пожалуйста решить)заранее спасибо=)

19)
$\int{\sin^5x\cos x}dx=\int{\sin^5x}d(\sin x)=\left|\sin x=t;d(\sin x)=dt\right|=\\ =\int{t^5}dt=\frac{1}{5+1}\cdot t^{5+1}+C=\frac16\sin^6x+C;$

20)
$\int{\frac{dx}{\arcsin^3x\sqrt{1-x^2}}}=\left|d(\arcsin x=\frac{dx}{\sqrt{1-x^2}})\right|=\\ =\int{\frac{d(\arcsin x)}{\arcsin^3x}}=\left|\arcsin x=t;\ d(\arcsin x)=dt \right|=\\ =\int{\frac{dt}{t^3}}=\int{t^{-3}}dt=\frac{1}{-3+1}\cdot t^{-3+1}+C=-\frac12t^{-2}+C=\\ =-\frac{1}{2t^2}+C=-\frac{1}{2\arcsin^2x}+C$

21)
$\int{\frac{ctg^7x}{\sin^2x}}dx=\left|d(ctg x)=d\left(\frac{\cos x}{\sin x}\right)=\frac{\cos'x\sin x-\cos x\sin'x}{\sin^2x}dx=\right|\\ =\left|\frac{-\cos^2x-\sin^2x}{\sin^2x}dx=-\frac{dx}{\sin^2x}\ \ \frac{dx}{\sin^2x}=-d(ctgx)\right|=\\ =-\int{ctg^7x\ }d(ctgx)=\left|ctgx=t;\ d(ctgx)=dt\right|=\\ =-\int{t^7}dt=-\frac{1}{7+1}\cdot t^{7+1}+C=-\frac{t^8}{8}+C=-\frac18ctg^8x+C$

22)
$\int{(3x+4)}e^{3x}dx=\left|\int{udv}-u\cdot v-\int{vdu}\right|=\\ \left|u=(3x+4);\ \ \ du=3dx\right|\\ \left|dv=e^{3x}dx;\ \ v=\frac13e^{3x}\right|\\ =\frac13(3x+4)e^{3x}-\int{\frac13e^{3x}3dx}=(x+\frac43)e^3x-\int{e^{3x}}dx=\\ =(x+\frac43)e^{3x}-\frac13e^{3x}+C=(x+\frac43-\frac13)e^{3x}+C=\\ =(x+1)e^{3x}+C.$