If `I_(n)=int_(0)^(pi) e^(x)sin^(n)x " dx then " (I_(3))/(I_(1))` is equal to

Options

A. `(3)/(5)`
B. `(1)/(5)`
C. 1
D. `(2)/(5)`

Answer

A

Explanation

We have,
`I_(n)=underset(0)overset(pi)int underset(II)e^(x)sin_(I)^(n)x dx `
`rArr I_(n)=[sin^(n)x e^(x)]_(0)^(x)-n underset(0)overset(pi)int underset(II)e^(x)sin_(I)^(n-1)x cosx dx`
`rArr I_(n)=-n underset(0)overset(pi)int underset(II)e^(x)sin_(I)^(n-1)x cos x dx`
`rArr I_(n)=-n[sin^(n-1)x cos x e^(x)]_(0)^(pi)+n underset(0)overset(pi)int e^(x){(n-1) sin^(n-2)x cos^(2)x-sin^(n)x}dx`
`rArr I_(n)=n(n-1)underset(0)overset(pi)int e^(x) sin^(n-2)x cos^(2)x dx-n underset(0)overset(pi)int e^(x)sin^(n) x dx`
`rArr I_(n)=n(n-1)underset(0)overset(pi)int e^(x) sin^(n-2)x(1-sin^(2)x)dx-nI_(n)`
`rArr I_(n)=n(n-1)underset(0)overset(pi)int e^(x)sin^(n-2)x dx-n(n-1)underset(0)overset(pi)int e^(x)sin^(n)x dx-nI_(n)`
`rArr I_(n)=n(n-1)I_(n-2)-n(n-1)I_(n)-nI_(n)`
`rArr I_(n)=(n(n-1))/(n^(2)+1)I_(n-2)`
`rArr (I_(n))/(I_(n-2))=(n(n-1))/(n^(2)+1)`
`rArr (I_(3))/(I_(1))=(6)/(10)=(3)/(5)`