The function `y=f(x)` is the solution of the differential equation `(dy)/(dx)+(x y)/(x^2-1)=(x^4+2x)/(sqrt(1-x^2))` in `(-1,1)` satisfying `f(0)=0.` Then `int_((sqrt(3))/2)^((sqrt(3))/2)f(x)dx` is (a) `( b ) (c) (d)pi/( e )3( f ) (g)-( h )(( i )sqrt(( j )3( k ))( l ))/( m )2( n ) (o) (p)` (q) (b) `( r ) (s) (t)pi/( u )3( v ) (w)-( x )(( y )sqrt(( z )3( a a ))( b b ))/( c c )4( d d ) (ee) (ff)` (gg) (c) `( d ) (e) (f)pi/( g )6( h ) (i)-( j )(( k )sqrt(( l )3( m ))( n ))/( o )4( p ) (q) (r)` (s) (d) `( t ) (u) (v)pi/( w )6( x ) (y)-( z )(( a a )sqrt(( b b )3( c c ))( d d ))/( e e )2( f f ) (gg) (hh)` (ii)

Options

A. `pi/3-sqrt(3)/2`
B. `pi/3-sqrt(3)/4`
C. `pi/6-sqrt(3)/4`
D. `pi/6-sqrt(3)/2`

Answer

B

Explanation

`(dy)/(dx) + x/(x^(2)-1)y=(x^(4)+2x)/sqrt(1-x^(2))`
This is a linear differential equation.
`I.F. =e^(intx/(x^(2)-1)dx)`
`e^(1/2"ln "|x^(2)-1|)=sqrt(1-x^(2))` (`therefore x in (-1,1))`
Therefore, solution is:
`ysqrt(1-x^(2))=int(x(x^(3)+2))/sqrt(1-x^(2)).sqrt(1-x^(2))dx`
or `ysqrt(1-x^(2))=int(x^(4)+2x)dx=x^(5)/5+x^(2)+C`
Since, `f(0)=0,C=0`
`therefore f(x) sqrt(1-x^(2))=x^(5)/5+x^(2)`
`rArr f(x) = x^(5)/(5sqrt(1-x^(2))+x^(2)/sqrt(1-x^(2))`
`therefore int_(-sqrt(3//2))^(sqrt(3)/2) f(x)dx= int_(sqrt(3)/2)^(sqrt(3)/2) (x^(5)/(5sqrt(1-x^(2))+x^(2)/sqrt(1-x^(2))))dx`
`=2int_(0)^(sqrt(3)//2)(x^(2)/sqrt(1-x^(2)))dx`
`rArr 2int_(0)^(pi//3) sin^(2)theta(d)theta`
`=2int_(0)^(pi//3)(1-cos2theta)/(2)(d)theta`
`=2[theta/2-(sin2theta)/4]_(0)^(pi//3)`
`=2(pi/6) -2(sqrt(3))/8 = pi/3-sqrt(3)/4`