Question

помогите пожалуйста решить)заранее спасибо=)

Answer 1

19)
$int{sin^5xcos x}dx=int{sin^5x}d(sin x)=left|sin x=t;d(sin x)=dtright|=\ =int{t^5}dt=frac{1}{5+1}cdot t^{5+1}+C=frac16sin^6x+C;$


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


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


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