Mathematics

If $$\displaystyle \int_{0}^{a} \dfrac{dx}{\sqrt{x+a}+\sqrt{x}}=\displaystyle \int_{0}^{\pi/8} \dfrac{2 \tan \theta}{\sin 2 \theta} d\theta$$, then value of $$a$$ is equal to $$(a > 0)$$


ANSWER

$$\dfrac{9}{16}$$


SOLUTION

$$\\LHS=\int_{0}^{a}(\dfrac{1}{\sqrt{x+a}+\sqrt{x}})dx\\=\int_{0}^{a}(\dfrac{1}{\sqrt{x+a}+\sqrt{x}})\times(\dfrac{\sqrt{x+a}-\sqrt{x}}{\sqrt{x+a}-\sqrt{x}})dx\\=\int_{0}^{a}(\dfrac{\sqrt{x+a}-\sqrt{x}}{(x+a)-x})dx$$

$$\\=(\dfrac{1}{a})\left[\int_{0}^{a}\sqrt{x+a}dx-\int_{0}^{a}\sqrt{x}dx\right]\\=(\dfrac{1}{a})\left[(\dfrac{(x+a)^{(\frac{3}{2})}}{(\dfrac{3}{2})})\right]_{0}^{a}-(\dfrac{1}{a})\left[(\dfrac{x^{(\frac{3}{2})}}{(\dfrac{3}{2})})\right]_{0}^{a}$$

$$\\=(\dfrac{1}{a})\left[(2a)^{(\frac{3}{2})}-a^{(\frac{3}{2})}\right]-(\dfrac{2}{3a})\left[a^{(\frac{3}{2})}-0\right]\\=(\dfrac{2}{3a})\>a^{(\frac{3}{2})}[2\sqrt{2}-1-1]\\=(\dfrac{2}{3a})\>a^{(\frac{3}{2})}(2\sqrt{2}-2)\\=(\dfrac{4}{3a})(\sqrt{2}-1)\>a^{(\frac{3}{2})}$$

$$\\=(\dfrac{4}{3})(\sqrt{2}-1)\>a^{(\frac{1}{2})}$$

$$\\RHS=\int_{0}^{(\frac{\pi}{8})}(\dfrac{2tan\theta}{sin2\theta})d\theta\\=\int_{0}^{(\frac{\pi}{8})}(\dfrac{2sin\theta}{cos\theta\times2sin\theta\>cos\theta})d\theta\\=\int_{0}^{(\frac{\pi}{8})}sec^2\theta\>d\theta=[tan\theta]_{0}^{\pi/8}\\=tan(\dfrac{\pi}{8})=(\sqrt{2}-1)\\=as\>LHS=RHS\\\therefore\>(\dfrac{4}{3})(\sqrt{2}-1)a^{(\frac{1}{2})}=(\sqrt{2}-1)\\\therefore\>a=(\dfrac{9}{16})$$


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Single Correct Medium Published on 17th 09, 2020
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