Which is larger: I_n = integral from 0 to (pi/2) sin^n x

Which is larger: $$ I_n = \int_0^{\frac{\pi}{2}} \sin^n x \, dx \quad \text{or} \quad I_{n+1} = \int_0^{\frac{\pi}{2}} \sin^{n+1} x \, dx, \quad (n \in \mathbb{N}) $$

Method: Direct comparison

Consider the difference:

$$I_n - I_{n+1} = \int_0^{\frac{\pi}{2}} \sin^n x \, dx - \int_0^{\frac{\pi}{2}} \sin^{n+1} x \, dx$$

$$= \int_0^{\frac{\pi}{2}} (\sin^n x - \sin^{n+1} x) \, dx = \int_0^{\frac{\pi}{2}} \sin^n x(1 - \sin x) \, dx$$

Step 1: Analyze $\sin x$ on $\left[0, \frac{\pi}{2}\right]$

For $x \in \left[0, \frac{\pi}{2}\right]$:

$$0 \leq \sin x \leq 1$$

Specifically:

• At $x = 0$: $\sin 0 = 0$
• At $x = \frac{\pi}{2}$: $\sin \frac{\pi}{2} = 1$
• For $x \in \left(0, \frac{\pi}{2}\right)$: $0 < \sin x < 1$

Step 2: Analyze $1 - \sin x$

For $x \in \left[0, \frac{\pi}{2}\right]$:

$$1 - \sin x \geq 0$$

Moreover, for $x \in \left(0, \frac{\pi}{2}\right)$: $1 - \sin x > 0$

Step 3: Analyze $\sin^n x$

For $x \in \left(0, \frac{\pi}{2}\right)$ and $n \geq 1$:

$$\sin^n x > 0$$

Step 4: Determine the sign of the integrand

For $x \in \left(0, \frac{\pi}{2}\right)$:

$$\sin^n x(1 - \sin x) > 0$$

At the endpoints:

• At $x = 0$: $\sin^n 0 \cdot (1 - 0) = 0 \cdot 1 = 0$
• At $x = \frac{\pi}{2}$: $\sin^n \frac{\pi}{2} \cdot (1 - 1) = 1 \cdot 0 = 0$

Therefore:

$$I_n - I_{n+1} = \int_0^{\frac{\pi}{2}} \sin^n x(1 - \sin x) \, dx > 0$$

The integrand is non-negative on $\left[0, \frac{\pi}{2}\right]$ and strictly positive on $\left(0, \frac{\pi}{2}\right)$, so the integral is positive.

$$\boxed{I_n > I_{n+1}}$$

$$\boxed{\int_0^{\frac{\pi}{2}} \sin^n x \, dx > \int_0^{\frac{\pi}{2}} \sin^{n+1} x \, dx}$$

Alternative explanation

For $x \in \left(0, \frac{\pi}{2}\right)$:

$$0 < \sin x < 1$$

Therefore, multiplying by $\sin x$ decreases the value:

$$\sin^{n+1} x = \sin^n x \cdot \sin x < \sin^n x \cdot 1 = \sin^n x$$

Integrating over $\left[0, \frac{\pi}{2}\right]$:

$$\int_0^{\frac{\pi}{2}} \sin^{n+1} x \, dx < \int_0^{\frac{\pi}{2}} \sin^n x \, dx$$

The sequence is strictly decreasing

This shows that the sequence of integrals:

$$I_0, I_1, I_2, I_3, \ldots$$

is strictly decreasing:

$$I_0 > I_1 > I_2 > I_3 > \cdots > I_n > I_{n+1} > \cdots$$

Verification with Wallis integrals:

The Wallis integrals give explicit formulas:

For even $n = 2m$:

$$I_{2m} = \int_0^{\frac{\pi}{2}} \sin^{2m} x \, dx = \frac{(2m-1)!!}{(2m)!!} \cdot \frac{\pi}{2}$$

For odd $n = 2m+1$:

$$I_{2m+1} = \int_0^{\frac{\pi}{2}} \sin^{2m+1} x \, dx = \frac{(2m)!!}{(2m+1)!!}$$

Numerical examples

$n = 0$:

$$I_0 = \int_0^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2} \approx 1.571$$

$n = 1$:

$$I_1 = \int_0^{\frac{\pi}{2}} \sin x \, dx = [-\cos x]_0^{\pi/2} = 0 - (-1) = 1$$

$n = 2$:

$$I_2 = \int_0^{\frac{\pi}{2}} \sin^2 x \, dx = \frac{1 \cdot \pi}{2 \cdot 2} = \frac{\pi}{4} \approx 0.785$$

$n = 3$:

$$I_3 = \int_0^{\frac{\pi}{2}} \sin^3 x \, dx = \frac{2}{3} \approx 0.667$$

$n = 4$:

$$I_4 = \int_0^{\frac{\pi}{2}} \sin^4 x \, dx = \frac{3 \cdot \pi}{4 \cdot 8} = \frac{3\pi}{16} \approx 0.589$$

Indeed: $I_0 > I_1 > I_2 > I_3 > I_4$ ✓

$$1.571 > 1.000 > 0.785 > 0.667 > 0.589$$

General principle

For any interval where $0 \leq f(x) \leq 1$:

$$\int f^{n+1}(x) \, dx < \int f^n(x) \, dx$$

This is because raising a number in $[0,1)$ to a higher power makes it smaller.

Summary

Comparison:

For $x \in \left(0, \frac{\pi}{2}\right)$:

$$0 < \sin x < 1 \implies \sin^{n+1} x < \sin^n x$$

Therefore:

$$\boxed{\int_0^{\frac{\pi}{2}} \sin^n x \, dx > \int_0^{\frac{\pi}{2}} \sin^{n+1} x \, dx}$$

Answer: $I_n$ is larger for all $n \in \mathbb{N}$.

Key insights:

• For $0 < t < 1$: higher powers are smaller ($t^{n+1} < t^n$)
• On $\left(0, \frac{\pi}{2}\right)$: $0 < \sin x < 1$ (except at endpoints)
• The sequence $\{I_n\}$ is strictly decreasing
• $I_n - I_{n+1} = \int_0^{\pi/2} \sin^n x(1-\sin x) dx > 0$
• Each integral is smaller than the previous one
• This result is fundamental in analyzing Wallis integrals