If `I_n=int(sinx)^ndx, n in N`, then `5I_4-6I_6` is equal to

Options

A. `sin x . (cos x )^(5) +C`
B. `sin 2x cos 2x +C`
C. `(sin 2x)/(8) [1+ cos ^(2) 2x -2 cos 2x ] +C`
D. `(sin 2x )/(8) [1+ cos ^(2) 2x + 2 cos 2x] +C`

Answer

To solve the problem \( 5I_4 - 6I_6 \), where \( I_n = \int \sin^n x \, dx \), we will use the reduction formula for the integral of sine raised to a power.

### Step-by-Step Solution:

1. **Reduction Formula**: The reduction formula for \( I_n = \int \sin^n x \, dx \) is given by:
\[ I_n = -\frac{1}{n} \sin^{n-1} x \cos x + \frac{n-1}{n} I_{n-2} \]

2. **Calculate \( I_4 \)**:
- Using the reduction formula for \( n = 4 \):
\[ I_4 = -\frac{1}{4} \sin^3 x \cos x + \frac{3}{4} I_2 \]

3. **Calculate \( I_2 \)**:
- Using the reduction formula for \( n = 2 \):
\[ I_2 = -\frac{1}{2} \sin x \cos x + \frac{1}{2} I_0 \]
- Since \( I_0 = \int dx = x + C \), we have:
\[ I_2 = -\frac{1}{2} \sin x \cos x + \frac{1}{2} (x + C) \]

4. **Substituting \( I_2 \) back into \( I_4 \)**:
- Substitute \( I_2 \) into the equation for \( I_4 \):
\[ I_4 = -\frac{1}{4} \sin^3 x \cos x + \frac{3}{4} \left(-\frac{1}{2} \sin x \cos x + \frac{1}{2} (x + C)\right) \]
- Simplifying this gives:
\[ I_4 = -\frac{1}{4} \sin^3 x \cos x - \frac{3}{8} \sin x \cos x + \frac{3}{8} (x + C) \]

5. **Calculate \( I_6 \)**:
- Using the reduction formula for \( n = 6 \):
\[ I_6 = -\frac{1}{6} \sin^5 x \cos x + \frac{5}{6} I_4 \]

6. **Substituting \( I_4 \) back into \( I_6 \)**:
- Substitute \( I_4 \) into the equation for \( I_6 \):
\[ I_6 = -\frac{1}{6} \sin^5 x \cos x + \frac{5}{6} \left(-\frac{1}{4} \sin^3 x \cos x - \frac{3}{8} \sin x \cos x + \frac{3}{8} (x + C)\right) \]

7. **Combine \( 5I_4 - 6I_6 \)**:
- Now, we compute \( 5I_4 - 6I_6 \):
\[ 5I_4 = 5\left(-\frac{1}{4} \sin^3 x \cos x - \frac{3}{8} \sin x \cos x + \frac{3}{8} (x + C)\right) \]
- And for \( 6I_6 \):
\[ 6I_6 = 6\left(-\frac{1}{6} \sin^5 x \cos x + \frac{5}{6} I_4\right) \]
- Combine these two to find \( 5I_4 - 6I_6 \).

8. **Final Simplification**:
- After simplification, we find that:
\[ 5I_4 - 6I_6 = \sin^5 x \cos x + C \]

### Final Answer:
Thus, \( 5I_4 - 6I_6 = \sin^5 x \cos x + C \).