∫tan4(1/2z)dz
let u=1/2z⇒du=1/2dzor2du=dz
؞∫tan4(1/2z)dz=2∫tan4udu…(i)
we define
ǀn=∫tannudu=(tann−1u)/(n−1)−ǀ(n−1)forn≥2
as the reduction formula
thus,
ǀ4=∫tan4udu=tan3u/3−ǀ2…(ii)
ǀ2=∫tan2udu=tanu−ǀ0…(iii)
ǀ0=∫du=u…(iv)
replacing (iv) in (iii) gives
ǀ(2)=tanu−u…(v)
replacing (v) in (ii) gives
ǀ4=tan3u/3−(tanu−u)
tan3u/3−tanu−u…(vi)
replacing (vi) in (i) gives
∫tan4(1/2z)dz=2[tan3(1/2z)/3−tan(1/2z)+1/2z]+c
=2/3tan3(1/2z)−2tan(1/2z)+z+c
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