Question

_____________________??

Answer 1

$\huge\boxed{\underline{\underline{\green{\tt Solution}}}}$

SUPPOSE

$\displaystyle \: I \: = \displaystyle \int\ { \frac{2x \: {sin}^{ - 1} 2x}{ \sqrt{1 - 4 {x}^{2} } } } \, dx ........(1)$

Let

$2x = sin \theta \:$

Then

$2 \: dx = cos \theta \: d\theta$

So From Equation (1)

$\displaystyle \: I \: = \displaystyle(1/2) \int\ { \frac{ \theta \: sin \theta \: }{ \sqrt{1 - {sin}^{2} \theta} } } \, cos d \theta$

$\displaystyle \: I \: = \displaystyle (1/2)\int\ { \frac{ \theta \: sin \theta \: }{ \sqrt{{cos}^{2} \theta} } } \, cos d \theta$

$\displaystyle \: I \: = \displaystyle (1/2)\int\ { \theta \: sin \theta \: } d \theta$

$\displaystyle \: I \: = \displaystyle (1/2)( - \theta \: cos \theta + sin \theta + c)$

Since

$2x = sin \theta \:$

So

$x \: = 0 \: gives \: \theta = 0$

$x = \frac{ \sqrt{3} }{4} \: \: gives \: \theta = \frac{\pi}{3}$

Hence the given Integral

$= \displaystyle \left. {(1/2)( - \theta cos \theta \: + sin \theta} + c) \right|_0^ \frac{\pi}{3}$

$= \displaystyle \:(1/2) ( - \frac{\pi}{3} cos \frac{\pi}{3} + \sin \frac{\pi}{3} + c) -(1/2) (0 + sin0 + c)$

$= \displaystyle \: ( - \frac{\pi}{12} + \frac{ \sqrt{3} }{4} )$

Answer 2

$\int _0^{\frac{1}{4}\sqrt{3}}\frac{\left(\:2xsin^{-1}2x\right)}{\sqrt{1-4x^2}}dx$

$ANSWER :-$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=2\cdot \int _0^{\frac{1}{4}\sqrt{3}}\frac{x\arcsin \left(2x\right)}{\sqrt{1-4x^2}}dx$

$\mathrm{Apply\:u-substitution:}\:u=2x$

$=2\cdot \int _0^{\frac{\sqrt{3}}{2}}\frac{u\arcsin \left(u\right)}{4\sqrt{1-u^2}}du$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=2\cdot \frac{1}{4}\cdot \int _0^{\frac{\sqrt{3}}{2}}\frac{u\arcsin \left(u\right)}{\sqrt{1-u^2}}du$

$\frac{u\arcsin \left(u\right)}{\sqrt{1-u^2}}=\left(u\arcsin \left(u\right)\right)\frac{1}{\sqrt{1-u^2}}$

$=2\cdot \frac{1}{4}\cdot \int _0^{\frac{\sqrt{3}}{2}}u\arcsin \left(u\right)\frac{1}{\sqrt{1-u^2}}du$

$\mathrm{Apply\:Integration\:By\:Parts:}\:u=\arcsin \left(u\right),\:v'=\frac{1}{\sqrt{1-u^2}}u$

$=2\cdot \frac{1}{4}\left[-\arcsin \left(u\right)\sqrt{1-u^2}-\int \:-1du\right]^{\frac{\sqrt{3}}{2}}_0$

$\int \:-1du=-u$

$= 2\cdot \frac{1}{4}\left[-\arcsin \left(u\right)\sqrt{1-u^2}-\left(-u\right)\right]^{\frac{\sqrt{3}}{2}}_0$

$\mathrm{Simplify\:}2\cdot \frac{1}{4}\left[-\arcsin \left(u\right)\sqrt{1-u^2}-\left(-u\right)\right]^{\frac{\sqrt{3}}{2}}_0:\quad \frac{1}{2}\left[-\arcsin \left(u\right)\sqrt{1-u^2}+u\right]^{\frac{\sqrt{3}}{2}}_0$

$=\frac{1}{2}\left[-\arcsin \left(u\right)\sqrt{1-u^2}+u\right]^{\frac{\sqrt{3}}{2}}_0$

$\mathrm{Compute\:the\:boundaries}:\quad \left[-\arcsin \left(u\right)\sqrt{1-u^2}+u\right]^{\frac{\sqrt{3}}{2}}_0=\frac{\sqrt{3}}{2}-\frac{\pi }{6}$

$=\frac{1}{2}\left(\frac{\sqrt{3}}{2}-\frac{\pi }{6}\right)$

$\mathrm{Simplify}$

$=\frac{3\sqrt{3}-\pi }{12}$