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Question Number 167024 by peter frank last updated on 04/Mar/22

∫sec θtan^4 θdθ

secθtan4θdθ

Answered by cortano1 last updated on 05/Mar/22

 let t = tan θ  I=∫ (t^4 /( (√(1+t^2 )))) = ∫ ((t^3 (t))/( (√(1+t^2 )))) dt  IBP  { ((u=t^3 ⇒du=3t^2  dt)),((v=∫(1/2) ((d(1+t^2 ))/( (√(1+t^2 )))) = (√(1+t^2 )))) :}  I=t^3 (√(1+t^2 ))−∫ 3t^2  (√(1+t^2 )) dt  I=t^3  (√(1+t^2 ))−3∫t (t(√(1+t^2 ))) dt  I=t^3  (√(1+t^2 )) −3[ (1/3)t (√((1+t^2 )^3 ))−(1/3)∫(1+t^2 )^(3/2) dt ]  I= t^3  (√(1+t^2 ))−t (√((1+t^2 )^3 ))+∫ (1+t^2 )^(3/2)  dt

lett=tanθI=t41+t2=t3(t)1+t2dtIBP{u=t3du=3t2dtv=12d(1+t2)1+t2=1+t2I=t31+t23t21+t2dtI=t31+t23t(t1+t2)dtI=t31+t23[13t(1+t2)313(1+t2)32dt]I=t31+t2t(1+t2)3+(1+t2)32dt

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