If `I_(n)=int(sinx+cosx)^(n)`dx, snd `I_(n)=1/n(sinx+cosx)^(n-1)(sinx-cosx)+(2k)/(n) I_(n-2)` then k=

Options

A. (n+1)
B. `(n-1)`
C. `(2n+1)`
D. `(2n-1)`

Answer

To solve the problem, we need to find the value of \( k \) in the given expression for \( I_n \). Let's break down the solution step by step.

### Step 1: Understand the given expression
We have:
\[ I_n = \int (\sin x + \cos x)^n \, dx \]
and the recurrence relation:
\[ I_n = \frac{1}{n} (\sin x + \cos x)^{n-1} (\sin x - \cos x) + \frac{2k}{n} I_{n-2} \]

### Step 2: Evaluate \( I_2 \)
Let’s first evaluate \( I_2 \):
\[ I_2 = \int (\sin x + \cos x)^2 \, dx \]
Expanding the integrand:
\[ (\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2 \sin x \cos x = 1 + \sin(2x) \]
Thus,
\[ I_2 = \int (1 + \sin(2x)) \, dx = x - \frac{1}{2} \cos(2x) + C \]

### Step 3: Substitute \( n = 2 \) in the recurrence relation
Now, substituting \( n = 2 \) in the recurrence relation:
\[ I_2 = \frac{1}{2} (\sin x + \cos x)^{1} (\sin x - \cos x) + \frac{2k}{2} I_0 \]
This simplifies to:
\[ I_2 = \frac{1}{2} (\sin x + \cos x)(\sin x - \cos x) + k I_0 \]

### Step 4: Evaluate \( I_0 \)
Now, we calculate \( I_0 \):
\[ I_0 = \int (\sin x + \cos x)^0 \, dx = \int 1 \, dx = x + C \]

### Step 5: Substitute \( I_0 \) into the equation
Substituting \( I_0 \) into the equation for \( I_2 \):
\[ I_2 = \frac{1}{2} (\sin x + \cos x)(\sin x - \cos x) + k(x + C) \]

### Step 6: Simplify the expression
The expression \( \frac{1}{2} (\sin x + \cos x)(\sin x - \cos x) \) can be simplified:
\[ \frac{1}{2} (\sin^2 x - \cos^2 x) = \frac{1}{2} \cdot \frac{1}{2} \sin(2x) = \frac{1}{2} \cdot \frac{1}{2} \cdot \sin(2x) \]
Thus, we have:
\[ I_2 = \frac{1}{2} \sin(2x) + kx + C \]

### Step 7: Compare coefficients
Now we compare the two expressions for \( I_2 \):
1. From direct integration: \( I_2 = x - \frac{1}{2} \cos(2x) + C \)
2. From recurrence: \( I_2 = \frac{1}{2} \sin(2x) + kx + C \)

### Step 8: Equate coefficients
Comparing the coefficients of \( x \):
\[ 1 = k \implies k = 1 \]

### Conclusion
Thus, the value of \( k \) is:
\[ \boxed{1} \]