Integrate In=∫sinnxdx by parts, splitting off one factor of sinx and using cos2x=1−sin2x to relate In to In−2.
Let In=∫sinnxdx=∫sinn−1x⋅sinxdx.
Integrate by parts with u=sinn−1x (du=(n−1)sinn−2xcosxdx) and dv=sinxdx (v=−cosx):
In=−sinn−1xcosx+(n−1)∫sinn−2xcos2xdx.
Using cos2x=1−sin2x:
In=−sinn−1xcosx+(n−1)∫sinn−2xdx−(n−1)∫sinnxdx=−sinn−1xcosx+(n−1)In−2−(n−1)In.
Collecting In terms:
In+(n−1)In=−sinn−1xcosx+(n−1)In−2⟹nIn=−sinn−1xcosx+(n−1)In−2.
In=−nsinn−1xcosx+nn−1In−2
Deducing ∫sin4xdx (i.e. I4):
I4=−4sin3xcosx+43I2.
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