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_(6","2)` is

Options

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

Answer

To establish the relation between \( I_{4,2} \) and \( I_{6,2} \), 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 \( m = 2 \). Thus:
\[ P(x) = \sin^{5} x \cos^{1} x \]

### Step 2: Differentiate \( P(x) \)
Now we differentiate \( P(x) \):
\[ \frac{dP}{dx} = (n+1) \sin^n x \cos^{m-1} x \cdot \cos x - (m-1) \sin^{n+1} x \cdot \sin x \cdot \cos^{m-2} x \]
Substituting \( n = 4 \) and \( m = 2 \):
\[ \frac{dP}{dx} = 5 \sin^4 x \cos x - 1 \cdot \sin^5 x \]

### Step 3: Simplify the differentiation
This simplifies to:
\[ \frac{dP}{dx} = 5 \sin^4 x \cos x - \sin^5 x \]

### Step 4: Rearranging the equation
We can rearrange the equation:
\[ \frac{dP}{dx} = 5 \sin^4 x \cos x - \sin^5 x = 5 \sin^4 x \cos x - \sin^5 x \]

### Step 5: Integrate both sides
Now we integrate both sides:
\[ P(x) = \int \left( 5 \sin^4 x \cos x - \sin^5 x \right) \, dx \]

### Step 6: Substitute back the integrals
Recognizing the integrals, we have:
\[ P(x) = 5 I_{4,2} - I_{6,2} \]

### Step 7: Solve for \( I_{4,2} \)
Thus, we can express \( I_{4,2} \) in terms of \( I_{6,2} \):
\[ I_{4,2} = \frac{1}{5} (P(x) + I_{6,2}) \]

### Final Relation
This gives us the relation between \( I_{4,2} \) and \( I_{6,2} \):
\[ I_{4,2} = \frac{1}{5} I_{6,2} + \text{constant} \]