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

Options

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

Answer

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

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

### Step 1: Define \( P(x) \)
We define \( P(x) \) for our specific case:

\[ P(x) = \sin^{4} x \cos^{2} x \]

### Step 2: Differentiate \( P(x) \)
Next, we differentiate \( P(x) \) using the product rule:

\[ \frac{dP}{dx} = \frac{d}{dx} (\sin^4 x) \cos^2 x + \sin^4 x \frac{d}{dx} (\cos^2 x) \]

Calculating each derivative:

1. \( \frac{d}{dx} (\sin^4 x) = 4 \sin^3 x \cos x \)
2. \( \frac{d}{dx} (\cos^2 x) = -2 \sin x \cos x \)

So we have:

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

### Step 3: Simplify the Derivative
Factoring out \( \cos x \):

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

### Step 4: Rearranging the Terms
Now, we can rearrange the terms:

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

### Step 5: Substitute \( I_{m,n} \) and \( I_{m-2,n} \)
Now we relate this to our integrals:

\[ \frac{dP}{dx} = 2 \cos x \left( 2 \sin^3 x \cos^2 x - \sin^4 x \right) = 2 \cos x \left( 2 \sin^3 x \cos^2 x - \sin^2 x \sin^2 x \right) \]

### Step 6: Integrate Both Sides
Integrating both sides gives:

\[ P(x) = 2 \int \cos x \left( 2 \sin^3 x \cos^2 x - \sin^4 x \right) dx \]

This can be expressed in terms of \( I_{4,2} \) and \( I_{2,2} \):

\[ I_{4,2} = 2I_{2,2} - \text{constant} \]

### Step 7: Final Relation
Thus, we can express the final relation as:

\[ I_{4,2} = \frac{3}{6} I_{2,2} - \frac{P}{6} \]

Where \( P \) is the integral of \( \sin^3 x \cos^3 x \).