L lgc New member Joined Oct 15, 2014 Messages 1 Oct 15, 2014 #1 I know I need to use trigonometric integration to solve but I am unsure if I should substitute sinx or cosx and then where to go from there. int sin5xcos3x dx thank you! Last edited: Oct 15, 2014
I know I need to use trigonometric integration to solve but I am unsure if I should substitute sinx or cosx and then where to go from there. int sin5xcos3x dx thank you!
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Oct 15, 2014 #2 lgc said: I know I need to use trigonometric integration to solve but I am unsure if I should substitute sinx or cosx and then where to go from there. int sin5xcos3x dx Click to expand... \(\displaystyle \sin^5(x)\cos^3(x)=\\\sin^5(x)\cos^2(x)\cos(x)\\ \sin^5(x)(1-\sin^2(x))\cos(x)\\ (\sin^5(x)-\sin^7(x))\cos(x)\) Now you have (u^5-u^7)du
lgc said: I know I need to use trigonometric integration to solve but I am unsure if I should substitute sinx or cosx and then where to go from there. int sin5xcos3x dx Click to expand... \(\displaystyle \sin^5(x)\cos^3(x)=\\\sin^5(x)\cos^2(x)\cos(x)\\ \sin^5(x)(1-\sin^2(x))\cos(x)\\ (\sin^5(x)-\sin^7(x))\cos(x)\) Now you have (u^5-u^7)du
