If int frac{dx}{1+3 sin^2 x} = frac{1}{2} tan | vedclass

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(A) We have the integral $I = \int \frac{dx}{1+3 \sin^2 x}$.Dividing the numerator and denominator by $\cos^2 x$,we get:$I = \int \frac{\sec^2 x dx}{\

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$2 	an x$

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(A) We have the integral $I = \int \frac{dx}{1+3 \sin^2 x}$.Dividing the numerator and denominator by $\cos^2 x$,we get:$I = \int \frac{\sec^2 x dx}{\sec^2 x + 3 	an^2 x} = \int \frac{\sec^2 x dx}{1 + 	an^2 x + 3 	an^2 x} = \int \frac{\sec^2 x dx}{1 + 4 	an^2 x}$.Let $t = 	an x$,then $dt = \sec^2 x dx$.The integral becomes:$I = \int \frac{dt}{1 + 4t^2} = \int \frac{dt}{1 + (2t)^2}$.Using the formula $\int \frac{du}{a^2 + u^2} = \frac{1}{a} 	an^{-1}(\frac{u}{a}) + c$,where $u = 2t$ and $du = 2dt$:$I = \frac{1}{2} \int \frac{2dt}{1 + (2t)^2} = \frac{1}{2} 	an^{-1}(2t) + c$.Substituting $t = 	an x$ back,we get:$I = \frac{1}{2} 	an^{-1}(2 	an x) + c$.Comparing this with the given expression $\frac{1}{2} 	an^{-1}(f(x)) + c$,we find $f(x) = 2 	an x$.

If $\int \frac{dx}{1+3 \sin^2 x} = \frac{1}{2} \tan^{-1}(f(x)) + c$,where $c$ is a constant of integration,then $f(x)$ is equal to

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