I'm trying to calculate ∫cos5(x)dx with the reduction formula.
∫cos5(x)dx=15cos4(x)sin(x)+45∫cos3(x)dx
then
∫cos3(x)dx=13cos2(x)sin(x)+23∫cos(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:
Answer
Converting the cosines in your answer to sines via the pythagorean identity, cos2x=1−sin2x yields:
15cos4(x)sin(x)+45(13cos2(x)sin(x)+23sin(x))
=15(1−sin2(x))2sin(x)+45(13(1−sin2(x))sin(x)+23sin(x))
Simplifying from there:
=15(1−2sin2(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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