If ∫$$1+$$$$sec$$⁡$$xdx=Ksin$$−$$1$$⁡$$(f(x))+C$$ then

Correct option is 1f(x)=2sin⁡(x/2),k=2

Questions

If $\int \sqrt{1 + \sec x} d x = K \sin^{- 1} (f (x)) + C$ then

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f(x)=2sin⁡(x/2),k=2

f(x)=2cos⁡(x/2),K=2

f(x)=2tan⁡(x/2),K=2

f(x)=2sin⁡(x/2),K=2

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detailed solution

Correct option is A

1+sec⁡x=1+cos⁡xcos⁡x=2cos⁡x/21−2sin2⁡x/2 soputting sin⁡x/2=t we have∫1+sec⁡xdx=∫2cos⁡x/21−2sin2⁡x/2dx=2sin−1⁡2t+C(t=sin⁡x/2)=2sin−1⁡(2sin⁡x/2)+Cso f(x)=2sin⁡x/2 and K=2

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If ∫1+sec⁡xdx=Ksin−1⁡(f(x))+C then

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If $\int \sqrt{1 + \sec x} d x = K \sin^{- 1} (f (x)) + C$ then