If intlimits_0^a {frac{{dx}}{{sqrt {x + a} + | vedclass

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(D) First,simplify the left-hand side $(LHS)$ integral by rationalizing the denominator:$\int_0^a \frac{1}{\sqrt{x+a} + \sqrt{x}} dx = \int_0^a \frac{

Vedclass · Accepted Answer

$\frac{9}{16}$

Vedclass · Answer

(D) First,simplify the left-hand side $(LHS)$ integral by rationalizing the denominator:$\int_0^a \frac{1}{\sqrt{x+a} + \sqrt{x}} dx = \int_0^a \frac{\sqrt{x+a} - \sqrt{x}}{(x+a) - x} dx = \frac{1}{a} \int_0^a (\sqrt{x+a} - \sqrt{x}) dx$$= \frac{1}{a} \left[ \frac{2}{3}(x+a)^{3/2} - \frac{2}{3}x^{3/2} ight]_0^a = \frac{2}{3a} [(2a)^{3/2} - a^{3/2} - (a^{3/2} - 0)] = \frac{2}{3a} [2\sqrt{2}a^{3/2} - 2a^{3/2}] = \frac{4\sqrt{a}}{3}(\sqrt{2}-1)$Now,simplify the right-hand side $(RHS)$ integral:$\int_0^{\pi/8} \frac{2	an 	heta}{\sin 2	heta} d	heta = \int_0^{\pi/8} \frac{2	an 	heta}{\frac{2	an 	heta}{1+	an^2 	heta}} d	heta = \int_0^{\pi/8} (1+	an^2 	heta) d	heta = \int_0^{\pi/8} \sec^2 	heta d	heta$$= [	an 	heta]_0^{\pi/8} = 	an(\pi/8) - 	an(0) = \sqrt{2}-1$Equating $LHS$ and $RHS$:$\frac{4\sqrt{a}}{3}(\sqrt{2}-1) = \sqrt{2}-1$$\frac{4\sqrt{a}}{3} = 1 \Rightarrow \sqrt{a} = \frac{3}{4} \Rightarrow a = \frac{9}{16}$

If $\int\limits_0^a {\frac{{dx}}{{\sqrt {x + a} + \sqrt x }}} = \int\limits_0^{\frac{\pi }{8}} {\frac{{2\tan \theta }}{{\sin 2\theta }}} d\theta$,then the value of $a$ is equal to $(a > 0)$.

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