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. (a) `pi/3-sqrt3/2`
B. (b) `pi/3-sqrt3/4`
C. (c) `pi/6-sqrt3/4`
D. (d) `pi/3-sqrt3/2`

Explanation

PLAN (i) Solution of the differential equation ` dy/dx+Py=Q` is
`y cdot (IF)= int Qcdot (IF) dx +c`
`if = e^(int Pdx)`
(ii)` int_(-a)^a f(x)dx=2int_(0)^a f(x) dx, if f(-x)=f(x)`
Given differential equation
` dy/dx +x/(x^(2)_1)y=(x^(2)+2x)/sqrt(1-x^(2)`
This is a linear differential equation.
`IF=e^(intx/(x^(2)-1)dx) = e^(1/2lnabs(x^(2)-1))=sqrt(1-x^(2))`
`rArr solution is y sqrt(1-x^(2))=int (x(x^(3)+2))/sqrt(1-x^(2)). sqrt(1-x^(2)) dx`
or ` y sqrt(1-x^(2)) = int (x^(4)+2x) dx = x^(5)/5+x^(2)+c`
`f(0)= 0 rArr c = 0 rArr f(x)sqrt(1-x^(2)) = x^(5)/5+x^(2)`
Now, `int_(-sqrt3//2)^(sqrt3//2) int (x) dx = int_(-sqrt3//2)^(sqrt3//2) x^(2)/(sqrt(1-x^(2)))dx`
[using property]
`= 2int_(0)^(sqrt3//2) x^(2)/(sqrt(1-x^(2)))dx`
`=2int_(0)^(pi//3) (sin^(2)theta)/cos theta cos theta d theta [taking x=sin theta]`
`=2int_(0)^(pi//3)sin^(2) theta d theta=int_(0)^(pi//3)(1- cos2 theta) d theta`
`=(theta-(sin2theta)/2)_(0)^(pi//3) =pi/3-(sin2pi//3)/2=pi/3-sqrt3/4`