作业帮∫ arccos7x dx∫ xcos(2-x) dx∫ sinx/(5+3sinx) dx
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/03 07:52:28
∫ arccos7x dx∫ xcos(2-x) dx∫ sinx/(5+3sinx) dx
∫ arccos7x dx
∫ xcos(2-x) dx
∫ sinx/(5+3sinx) dx
∫ arccos7x dx∫ xcos(2-x) dx∫ sinx/(5+3sinx) dx
∫ arccos7x dx
分部积分法
∫ arccos7x dx=x*arccos7x-∫ xdarccos7x=x*arccos7x+∫ 7x / √ 1-(7x)^2 dx
∫ 7x / √ 1-(7x)^2 dx=∫ 1/14* (1-(7x)^2)^-1/2 d(7x)^2=1/7*√ 1-(7x)^2 +C
∴ =x*arccos7x+1/7*√[ 1-(7x)^2]+C
∫ xcos(2-x) dx
分部积分法
∫ xcos(2-x) dx=∫ xcos(x-2) dx=sin(x-2)*x-∫ sin(x-2) dx=sin(x-2)*x+cos(x-2) +C
∫ sinx/(5+3sinx) dx
=∫[ 1/3(3sinx+5)-5/3]/(5+3sinx) dx=∫(1/3)-(5/3)/(5+3sinx) dx
∫1/(5+3sinx) dx 令t=tan(x/2) x=2atant dx=2/(1+t^2) dt sinx=2t/1+t^2
代入后即是
∫1/(5+3sinx) dx=∫2/(1+t^2)/(5+6t/1+t^2) dt=∫2/(5+5t^2+6t) dt=2/5∫1/(1+t^2+6/5 t) dt
=2/5∫1/(t^2+6/5 t+9/25+16/25) dt=2/5 ∫1/[(t+3/5)^2+16/25] dt=2/5 * 25/16 ∫1/[5/4*t+3/4]^2+1] dt
=1/2∫1/[5/4*t+3/4]^2+1] d[5/4*t+3/4]=1/2 * arctan(5/4*t+3/4)+C
∴=x/3 - 5/6 * arctan(5/4*tan(x/2)+3/4)+C