So I've been trying to compute ∫sin4(x)dx
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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