If f(x) = frac{1}{(cos^2 x) sqrt{1 + tan x}}, | vedclass

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(C) We are given $f(x) = \frac{1}{\cos^2 x \sqrt{1 + 	an x}} = \sec^2 x (1 + 	an x)^{-1/2}$.Let $t = 1 + 	an x$. Then $dt = \sec^2 x \ dx$.Substitu

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$2 (\sqrt{1 + 	an x} + 1)$

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(C) We are given $f(x) = \frac{1}{\cos^2 x \sqrt{1 + 	an x}} = \sec^2 x (1 + 	an x)^{-1/2}$.Let $t = 1 + 	an x$. Then $dt = \sec^2 x \ dx$.Substituting these into the integral:$F(x) = \int f(x) \ dx = \int (1 + 	an x)^{-1/2} \sec^2 x \ dx = \int t^{-1/2} \ dt$.Integrating with respect to $t$:$F(x) = \frac{t^{1/2}}{1/2} + C = 2 \sqrt{t} + C = 2 \sqrt{1 + 	an x} + C$.Given $F(0) = 4$,we substitute $x = 0$:$F(0) = 2 \sqrt{1 + 	an 0} + C = 2 \sqrt{1 + 0} + C = 2 + C$.Since $F(0) = 4$,we have $2 + C = 4$,which implies $C = 2$.Thus,$F(x) = 2 \sqrt{1 + 	an x} + 2 = 2 (\sqrt{1 + 	an x} + 1)$.

If $f(x) = \frac{1}{(\cos^2 x) \sqrt{1 + \tan x}}$,then its anti-derivative $F(x) = . . . . . . .$,given $F(0) = 4$.

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