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

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

Using integration by parts, we can write:

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

Let \( u = \cos^n x \) and \( dv = \cos(nx) \, dx \). Then, we have:

\[ du = -n \cos^{n-1} x \sin x \, dx, \quad v = \frac{\sin(nx)}{n}. \]

### Step 2: Apply integration by parts

Now, applying integration by parts:

\[ I_n = \left[ \cos^n x \cdot \frac{\sin(nx)}{n} \right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} \frac{\sin(nx)}{n} \cdot (-n \cos^{n-1} x \sin x) \, dx. \]

Evaluating the boundary term:

At \( x = 0 \): \( \cos^n(0) \cdot \frac{\sin(n \cdot 0)}{n} = 0 \).

At \( x = \frac{\pi}{2} \): \( \cos^n\left(\frac{\pi}{2}\right) \cdot \frac{\sin(n \cdot \frac{\pi}{2})}{n} = 0 \).

Thus, the boundary term evaluates to zero:

\[ I_n = 0 + \int_0^{\frac{\pi}{2}} \sin(nx) \cos^{n-1} x \sin x \, dx. \]

### Step 3: Simplify the integral

Now, we can simplify the integral:

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

Using the identity \( \sin x = \frac{1}{2}(\sin(2x)) \):

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

### Step 4: Relate \( I_n \) and \( I_{n+1} \)

We can express \( I_{n+1} \) similarly:

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

Using the same integration by parts technique, we can find a relationship between \( I_n \) and \( I_{n+1} \):

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

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

From the relationship derived above, we can express the ratio:

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

### Conclusion

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

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