Comprehension 1
Let `I_(n,m)=intsin^(n)xcos^(m)x.dx`. Then we can relate `I_(n,m)` with each of the following
i) `I_(n-2),m`, ii) `I_(n+2),m`, iii) `I_(n,m-2)`
iv) `I_(n,m-2)`, v) `I_(n-2,m+2)`, vi) `I_(n+2,m-2)`
Suppose we want to establish a relation between `I_(n,m)` and `I_(n,m-2)`, then we set
`P(x)=sin^(n+1)xcos^(m-1)x`................(1)
In `I_(n,m)` and `I_(n,m-2)` the exponent of `cosx` is m and `m-2+1=m-1`. Now choose the exponent `m-1` of `cosx` in P(x). Similarly choose hte exponent of `sinx` for P(x).
Now, differentiating both sides of (1), we get
`P^(')(x) = (n+1)sin^(n)xcos^(m)X-(m-1)sin^(n+2)Xcos^(m-2)X`
`=(n+1)sin^(n)Xcos^(m)X-(m-1)sin^(n)x(1-cos^(2)x)cos^(m-2)X`
`=(n+1)sin^(n)X cos^(m)X-(m-1)sin^(n)Xcos^(m-2)X+(m-1)sin^(n)Xcos^(m)X`
`=(n+m)sin^(n)Xcos^(m)X-(m-1)sin^(n)Xcos^(m-2)X`
Now, integrating both sides, we get
`sin^(n+1)cos^(m-1)x=(n+m)I_(n,m)-(m-1)I_(n,m+2)`
Similarly, we can establish the other relations.
The relation between `I_(4,2)` and `I_(2,2)` is

Options

A. `I_(4,2)=1/6 (-sin^(3)xcos^(3)x+3I_(2,2))`
B. `I_(4,2)=1/6(sin^(3)x cos^(3)x-3I_(2,2))`
C. `I_(4,2)=1/6(sin^(3)xcos^(3)x-3I_(2,2))`
D. `I_(4,2)=1/6(sin^(3)xcos^(3)x-3I_(2,2))`

Answer

`

Explanation

A