Find out the area bounded by the curve `y=int_(1//8)^(sin^(2)x)(sin^(-1)sqrt(t))dt+int_(1//8)^(cos^(2)x)(cos^(-1)sqrt(t))dt(0lexlepi//2)` and the curve satisfying the differential equation `y(x+y^3) dx=x(y^3-x)dy` passing through (`4,-2)`.

Options

A. `pi`
B. `(pi)/(2)`
C. `(pi)/(4)`
D. none of these

Answer

C

Explanation

Let `phi(x)=overset(sin^(2)x)underset(0)int sin^(-1)sqrt(t)dt+overset(cos^(2)x)underset(0)int cos^(-1)sqrt(t)dt`.
Then,
`(dphi)/(dx)=overset(sin^(2)x)underset(0)int 0dt+{(d)/(dx)(sin^(2)x)}xxsin^(-1)(sqrt(sin^(2)x))-0+overset(cos^(2))underset(0)int 0 dt+{(d)/(dx)(cos^(2)x)}xxcos^(-1)(sqrt(cos^(2)x))-0`
`rArr (dphi)/(dx)=(2 sin xcos x)x-(2 sin x cos x)x`
`rArr(dphi)/(dx)=0` for all x
`:.phi(x)=` Consider for all x
Let`phi(x)=k`, for all x `" "`......(i)
`rArr phi((pi)/(4))=k`
`rArr overset(1//2)underset(0)int sin^(-1)sqrt(t)dt+overset(1//2)underset(0)int cos^(-1)sqrt(t)dt=k`
`rArr overset(1//2)underset(0)int (sin^(-1))sqrt(t)+cos^(-1)sqrt(t)dt=k`
`rArr overset(1//2)underset(0)int (pi)/(2)dt=k rArr k=(pi)/(4)`
Putting `k=pi//4` in (i), we get `phi(x)=pi//4` for all x.