Question

помогите решить неопределенные интегралы:
1) $(1) intlimits dx/ sqrt{6x-x^2} (2) intlimits cos(^5)4xdx (3)intlimits xdx/x^2+10 4)intlimits dx/2x+3$

Answer 1

$1);intfrac{dx}{sqrt{6x-x^2}}=intfrac{dx}{sqrt{9-9+6x-x^2}}=intfrac{dx}{sqrt{9-(x-3)^2}}=left(begin{array}{c}u=x-3\du=dxend{array}right)=\=intfrac{du}{sqrt{9-u^2}}=intfrac{du}{sqrt{3^2-u^2}}=arcsinfrac{u}3+C=arcsinleft(frac{x-3}3right)+C\\2);intcos^54x;dx=left(begin{array}{c}u=4x\du=4dxend{array}right)=frac14intcos^54x;dx\intcos^nax;dx=frac{cos^{n-1}axcdotsin ax}{an}+frac{n-1}nintcos^{n-2}ax;dx$

$frac14intcos^54x;dx=frac14left(frac15sin ucos^4u+frac45intcos^3u;duright)=\=frac1{20}sin ucos^4u+frac15intcos^3u;du=frac1{20}sin ucos^4u+frac1{15}sin ucos^2u+\+frac2{15}intcos u;du=frac1{20}sin ucos^4u+frac1{15}sin ucos^2u+frac2{15}sin u+C=\=frac1{960}left(150sin4x+25sin12x+3sin20xright)+C\\3);intfrac{x;dx}{x^2+10}=left(begin{array}{c}u=x^2+10\du=2x;dxend{array}right)=frac12intfrac{du}u=frac12ln u+C=\=frac12ln(x^2+10)+C$

$4);intfrac{dx}{2x+3}=left(begin{array}{c}u=2x+3\du=2dxend{array}right)=frac12intfrac{du}u=frac12ln u+C=frac12ln(2x+3)+C$