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Let (1+x4)dx(1x4)32=f(x)+c1 where f(0) = 0 and f(x).dx=g(x)+c2 with g(0) = 0.If g(12)=π2k, then value of k is___


Solution

(1+x4)dx(1x4)32=(x+1x3)(1x2x2)32dx
Put 1x2x2=t2
 2(x+1x3)dx=2t dt
f(x)=tt3dt=1t+C
=x1x4+C1
as f(0)=0C1=0
Now, x1x4dx=12 sin1x2+C2
g(0)=0C2=0
g(x)=12 sin1(x2)
g(12)=π12

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