So I've been trying to compute ∫sin4(x)dx and everywhere they use the reduction formula which we haven't learned yet so I've been wondering if theres another way to do it? Thanks in advance.
Answer
Performing integration by parts,
∫x0sin2tdt=[−costsint]x0+∫x0cos2tdt=−cosxsinx+∫x0(1−sin2t)dt=−cosxsinx+∫x01dt−∫x0sin2tdt=−cosxsinx+x−∫x0sin2tdt
Therefore,
∫x0sin2tdt=−12cosxsinx+12x
∫x0sin4tdt=∫x0(1−cos2)sin2tdt=∫x0sin2tdt−∫x0cos2tsin2tdt=−12cosxsinx+12x−∫x0cos2tsin2tdt
Since, for t real,
sin(2t)=2sintcost
then,
∫x0sin4tdt=−12cosxsinx+12x−14∫x0sin2(2t)dt
In the latter integral perform the change of variable y=2t,
∫x0sin4tdt=−12cosxsinx+12x−18∫2x0sin2(y)dy=−12cosxsinx+12x−18(−12cos(2x)sin(2x)+12×2x)=−14sin(2x)+12x+132sin(4x)−18x=−14sin(2x)+38x+132sin(4x)
Therefore,
∫sin4xdx=38x+132sin(4x)−14sin(2x)+C
(C a real constant)
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