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 `I_(4","2)` and `I_(4","4)` is

Options

A. `I_(4","2)=1/3 (sin^(5)x cos^(3)x+8I_(4","4))`
B. `I_(4","2)=1/3 (- sin^(5)x cos^(3)x+8I_(4","4))`
C. `I_(4","2)=1/3 (sin^(5)x cos^(3)x- 8I_(4","4))`
D. `I_(4","2)=1/3 (sin^(5)x cos^(3)x+6I_(4","4))`

Answer

To establish the relation between \( I_{4,2} \) and \( I_{4,4} \), we start with the definitions of the integrals:

\[ I_{m,n} = \int \sin^n x \cos^m x \, dx \]

### Step 1: Define \( P(x) \)

We define:

\[ P(x) = \sin^{n+1} x \cos^{m-1} x \]

For our case, we have \( n = 4 \) and \( m = 4 \).

### Step 2: Differentiate \( P(x) \)

Now, we differentiate \( P(x) \):

\[ \frac{dP}{dx} = (n+1) \sin^n x \cos^{m-1} x \cos x - (m-1) \sin^{n+1} x \sin x \cos^{m-2} x \]

Substituting \( n = 4 \) and \( m = 4 \):

\[ \frac{dP}{dx} = 5 \sin^4 x \cos^3 x - 3 \sin^5 x \cos^2 x \]

### Step 3: Rewrite the Derivative

We can rewrite the derivative as:

\[ \frac{dP}{dx} = 5 \sin^4 x \cos^3 x - 3 \sin^5 x \cos^2 x \]

### Step 4: Integrate Both Sides

Integrating both sides gives us:

\[ P(x) = 5 I_{4,3} - 3 I_{5,2} \]

### Step 5: Substitute Back for \( P(x) \)

Substituting back for \( P(x) \):

\[ \sin^5 x \cos^3 x = 5 I_{4,3} - 3 I_{5,2} \]

### Step 6: Establish the Relation

Now, we need to relate \( I_{4,2} \) and \( I_{4,4} \). We can express \( I_{4,2} \) in terms of \( I_{4,4} \):

\[ I_{4,4} = \int \sin^4 x \cos^4 x \, dx \]
\[ I_{4,2} = \int \sin^4 x \cos^2 x \, dx \]

### Final Relation

From our previous steps, we can derive:

\[ I_{4,2} = \frac{1}{3} \left( 5 I_{4,4} - P(x) \right) \]

This gives us the relation between \( I_{4,2} \) and \( I_{4,4} \).