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Let I(m","n)= int sin^(n)x cos^(m)x dx. ...

Let `I_(m","n)= int sin^(n)x cos^(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)" " I_(n+2","m-2)`
Suppose we want to establish a relation between `I_(n","m)` and `I_(n","m-2)`, then we get
`P(x)=sin^(n+1)x cos^(m-1)x` ...(i)
In `I_(n","m)` and `I_(n","m-2)` the exponent of cos x in m and `m-2` respectively, the minimum of the two is m - 2, adding 1 to the minimum we get `m-2+1=m-1`. Now, choose the exponent of sin x for m - 1 of cos x in P(x). Similarly, choose the exponent of sin x for
`P(x)=(nH)sin^(n)x cos^(m)x-(m-1)sin^(n+2) x cos^(m-2)x`.
Now, differentiating both the sides of Eq. (i), we get
`=(n+1)sin^(n)x cos^(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)x cos^(m-2)x+(m-1)sin^(n)x cos^(n)x`
`=(n+m)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x`
Now, integrating both the sides, we get
`sin^(n+1)x 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

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))`

Text Solution

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The correct Answer is:
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A
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