If int frac{dx}{sqrt{16-9x^2}} = A sin^{-1}(B | vedclass

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(D) We know that $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(\frac{x}{a}) + C$. Given integral is $I = \int \frac{dx}{\sqrt{4^2 - (3x)^2}}$. Using t

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$\frac{13}{12}$

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(D) We know that $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(\frac{x}{a}) + C$. Given integral is $I = \int \frac{dx}{\sqrt{4^2 - (3x)^2}}$. Using the substitution $u = 3x$,we have $du = 3dx$,so $dx = \frac{du}{3}$. $I = \int \frac{du/3}{\sqrt{4^2 - u^2}} = \frac{1}{3} \int \frac{du}{\sqrt{4^2 - u^2}}$. $I = \frac{1}{3} \sin^{-1}(\frac{u}{4}) + C = \frac{1}{3} \sin^{-1}(\frac{3x}{4}) + C$. Comparing this with $A \sin^{-1}(Bx) + C$,we get $A = \frac{1}{3}$ and $B = \frac{3}{4}$. Therefore,$A + B = \frac{1}{3} + \frac{3}{4} = \frac{4 + 9}{12} = \frac{13}{12}$.

If $\int \frac{dx}{\sqrt{16-9x^2}} = A \sin^{-1}(Bx) + C$,then $A+B=$

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