?? R0- R certainly should "get rid of" the R variable. I'm not completely certain whether that first R0 is in the denominator with pi or not. I'll assume it's not.
What you have is [tex]\frac{20}{\pi}R_0(R_0)^{-\frac{1}{7}}(2\pi)\int(R_0-R)^{\frac{1}{7}}RdR[/tex][tex]=40R_0^{\frac{6}{7}}\int(R_0-R)^{\frac{1}{7}}RdR[/tex].
Let y= R_{0}- R so that dy= -dR and R= R_{0}- y. Then the integral becomes [tex]-40R_0^{\frac{6}{7}}\int y^{\frac{1}{7}}(R_0-y)dy= -40R_0^{\frac{6}{7}}\int(R_0y^{\frac{1}{7}}- y^{\frac{8}{7}})dy[/tex] which is easy.
We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling We Value Civility
• Positive and compassionate attitudes
• Patience while debating We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving