Wednesday, 3 August 2016

integration - intcos5(x)dx with reduction formula



I'm trying to calculate cos5(x)dx with the reduction formula.



cos5(x)dx=15cos4(x)sin(x)+45cos3(x)dx
then
cos3(x)dx=13cos2(x)sin(x)+23cos(x)dx

so
cos5(x)dx=15cos4(x)sin(x)+45(13cos2(x)sin(x)+23sin(x))



But this looks different from:



15sin5(x)23sin3(x)+sin(x)



given by:



https://www.symbolab.com/solver/step-by-step/%5Cint%20cos%5E%7B5%7D%5Cleft%28x%5Cright%29dx/?origin=button



Answer



Converting the cosines in your answer to sines via the pythagorean identity, cos2x=1sin2x yields:



15cos4(x)sin(x)+45(13cos2(x)sin(x)+23sin(x))
=15(1sin2(x))2sin(x)+45(13(1sin2(x))sin(x)+23sin(x))



Simplifying from there:



=15(12sin2(x)+sin4(x))sin(x)+45(13sin(x)13sin3(x))+23sin(x))




=15sin(x)25sin3(x)+15sin5(x)+415sin(x)415sin3(x))+815sin(x)



=15sin5(x)23sin3(x)+sin(x)



Which agrees with the answer from the solver.


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