Question
If Φ(x)=limn→∞xn−x−nxn+x−n,0<x<1,ϵN, then ∫(sin−1x)(Φ(x))dx is equal to
Answer: Option A
:
A
We have ϕ(x)=limn→∞x2n−1x2n+1=1,0<x<1∴∫sin−1x.ϕ(x)dx=∫sin−1xdx=[xsin−1x+√1−x2]+c
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A
We have ϕ(x)=limn→∞x2n−1x2n+1=1,0<x<1∴∫sin−1x.ϕ(x)dx=∫sin−1xdx=[xsin−1x+√1−x2]+c
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