Let `I_(n)=int_(0)^(pi//2) cos^(n)x cos nx dx`. Then, `I_(n):I_(n+1)` is equal to

Options

A. `3:1`
B. `2:3`
C. `2:1`
D. `3:4`

Answer

To solve the problem, we need to find the ratio \( I_n : I_{n+1} \) where

\[ I_n = \int_0^{\frac{\pi}{2}} \cos^n x \cos(nx) \, dx. \]

### Step 1: Express \( I_n \) using integration by parts

We can apply integration by parts to \( I_n \). We set:
- \( u = \cos^n x \)
- \( dv = \cos(nx) \, dx \)

Then, we differentiate and integrate:
- \( du = -n \cos^{n-1} x \sin x \, dx \)
- \( v = \frac{\sin(nx)}{n} \)

Using integration by parts, we have:

\[ I_n = \left[ u v \right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} v \, du \]

### Step 2: Evaluate the boundary term

Evaluating the boundary term \( \left[ u v \right]_0^{\frac{\pi}{2}} \):

At \( x = \frac{\pi}{2} \), \( \cos\left(\frac{\pi}{2}\right) = 0 \), so \( u v = 0 \).

At \( x = 0 \), \( \cos(0) = 1 \) and \( \sin(0) = 0 \), so \( u v = 0 \).

Thus, the boundary term is:

\[ \left[ u v \right]_0^{\frac{\pi}{2}} = 0 - 0 = 0. \]

### Step 3: Substitute into the integral

Now, substituting into the integral gives us:

\[ I_n = -\int_0^{\frac{\pi}{2}} \frac{\sin(nx)}{n} \left(-n \cos^{n-1} x \sin x \right) \, dx. \]

This simplifies to:

\[ I_n = \int_0^{\frac{\pi}{2}} \cos^{n-1} x \sin^2 x \sin(nx) \, dx. \]

### Step 4: Express \( I_{n+1} \)

Now, let's express \( I_{n+1} \):

\[ I_{n+1} = \int_0^{\frac{\pi}{2}} \cos^{n+1} x \cos(nx) \, dx. \]

### Step 5: Form the ratio \( \frac{I_n}{I_{n+1}} \)

Now we can find the ratio \( \frac{I_n}{I_{n+1}} \):

\[ \frac{I_n}{I_{n+1}} = \frac{\int_0^{\frac{\pi}{2}} \cos^{n-1} x \sin^2 x \sin(nx) \, dx}{\int_0^{\frac{\pi}{2}} \cos^{n+1} x \cos(nx) \, dx}. \]

### Step 6: Simplify the ratio

Notice that:

\[ I_n = \frac{n}{n+1} I_{n-1}. \]

From the recurrence relation, we can derive:

\[ \frac{I_n}{I_{n+1}} = \frac{n}{n+1}. \]

### Conclusion

Thus, the ratio \( I_n : I_{n+1} \) simplifies to:

\[ \frac{I_n}{I_{n+1}} = \frac{2}{1}. \]

Hence, the answer is:

\[ I_n : I_{n+1} = 2 : 1. \]