Question

If ${\int }_0^a\frac{dx}{\sqrt{x+a}+\sqrt{x}}={\int }_0^{\pi /8}\frac{2\tan \theta }{\sin 2\theta }d\theta$, then value of $a$ is equal to $(a>0)$

• A. $\frac{3}{4}$
• B. $\frac{\pi }{4}$
• C. $\frac{3\pi }{4}$
• D. $\frac{9}{16}$

Answer 1

The correct option is D $\frac{9}{16}$

$LHS={\int }_0^a\left(\frac{1}{\sqrt{x+a}+\sqrt{x}}\right)dx={\int }_0^a\left(\frac{1}{\sqrt{x+a}+\sqrt{x}}\right)\times \left(\frac{\sqrt{x+a}-\sqrt{x}}{\sqrt{x+a}-\sqrt{x}}\right)dx={\int }_0^a\left(\frac{\sqrt{x+a}-\sqrt{x}}{(x+a)-x}\right)dx$

$=\left(\frac{1}{a}\right)[{\int }_0^a\sqrt{x+a}dx-{\int }_0^a\sqrt{x}dx]=\left(\frac{1}{a}\right){\left[\right(\frac{(x+a{)}^{\left(\frac{3}{2}\right)}}{\left(\frac{3}{2}\right)}\left)\right]}_0^a-\left(\frac{1}{a}\right){\left[\right(\frac{x^{\left(\frac{3}{2}\right)}}{\left(\frac{3}{2}\right)}\left)\right]}_0^a$

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

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

$RHS={\int }_0^{\left(\frac{\pi }{8}\right)}\left(\frac{2tan\theta }{sin2\theta }\right)d\theta ={\int }_0^{\left(\frac{\pi }{8}\right)}\left(\frac{2sin\theta }{cos\theta \times 2sin\theta cos\theta }\right)d\theta ={\int }_0^{\left(\frac{\pi }{8}\right)}se{c}^2\theta d\theta =[tan\theta {]}_0^{\pi /8}=tan(\frac{\pi }{8})=(\sqrt{2}-1)=asLHS=RHS\therefore (\frac{4}{3}\left)\right(\sqrt{2}-1)a^{\left(\frac{1}{2}\right)}=(\sqrt{2}-1)\therefore a=(\frac{9}{16})$