Prove that integral from 0 to (pi/2) sin^n x dx = (n-1)/n

Prove that $\int_0^{\pi/2} \sin^n x \, dx = \frac{n-1}{n} \int_0^{\pi/2} \sin^{n-2} x \, dx$, and use this to find $\int_0^{\pi/2} \sin^6 x \, dx$.

Part 1: Prove the reduction formula

We will use integration by parts.

Step 1: Rewrite $\sin^n x$

\[\sin^n x = \sin^{n-1} x \cdot \sin x\]

Step 2: Apply integration by parts

Let $u = \sin^{n-1} x$, so $u' = (n-1)\sin^{n-2} x \cos x$; and $dv = \sin x \, dx$, so $v = -\cos x$.

\[\int_0^{\pi/2} \sin^n x \, dx = \left[-\cos x \sin^{n-1} x\right]_0^{\pi/2} + (n-1)\int_0^{\pi/2} \sin^{n-2} x \cos^2 x \, dx\]

Step 3: Evaluate the boundary term

\[\left[-\cos x \sin^{n-1} x\right]_0^{\pi/2} = -0 \cdot 1 + 1 \cdot 0 = 0\]

Step 4: Use the identity $\cos^2 x = 1 - \sin^2 x$

\[\int_0^{\pi/2} \sin^n x \, dx = (n-1)\int_0^{\pi/2} \sin^{n-2} x \, dx - (n-1)\int_0^{\pi/2} \sin^n x \, dx\]

Step 5: Solve for $\int_0^{\pi/2} \sin^n x \, dx$

Let $I_n = \int_0^{\pi/2} \sin^n x \, dx$. Then $I_n = (n-1)I_{n-2} - (n-1)I_n$, so $nI_n = (n-1)I_{n-2}$, giving:

\[\boxed{\int_0^{\pi/2} \sin^n x \, dx = \frac{n-1}{n} \int_0^{\pi/2} \sin^{n-2} x \, dx}\]

Part 2: Calculate $\int_0^{\pi/2} \sin^6 x \, dx$

We apply the reduction formula repeatedly.

Step 1: Find the base cases

\[I_0 = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2}\]

Using the reduction formula with $n=2$:

\[I_2 = \frac{1}{2} I_0 = \frac{\pi}{4}\]

Step 2: Calculate $I_4$

\[I_4 = \frac{3}{4} I_2 = \frac{3}{4} \cdot \frac{\pi}{4} = \frac{3\pi}{16}\]

Step 3: Calculate $I_6$

\[I_6 = \frac{5}{6} I_4 = \frac{5}{6} \cdot \frac{3\pi}{16} = \frac{15\pi}{96} = \frac{5\pi}{32}\]

Verification of the pattern:

• $I_0 = \frac{\pi}{2}$
• $I_2 = \frac{1\cdot\pi}{2\cdot2} = \frac{\pi}{4}$
• $I_4 = \frac{1\cdot3\cdot\pi}{2\cdot2\cdot4} = \frac{3\pi}{16}$
• $I_6 = \frac{1\cdot3\cdot5\cdot\pi}{2\cdot2\cdot4\cdot6} = \frac{5\pi}{32}$

General pattern:

For even $n=2m$:

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

where $(2m-1)!! = 1\cdot3\cdot5\cdots(2m-1)$ and $(2m)!! = 2\cdot4\cdot6\cdots(2m)$. For $m=3$ ($n=6$):

\[I_6 = \frac{5!!}{6!!} \cdot \frac{\pi}{2} = \frac{1\cdot3\cdot5}{2\cdot4\cdot6} \cdot \frac{\pi}{2} = \frac{5\pi}{32}\]

Numerical value:

\[\frac{5\pi}{32} \approx 0.491\]

Final answer:

\[\boxed{\int_0^{\pi/2} \sin^6 x \, dx = \frac{5\pi}{32}}\]

Note:

The reduction formula is also known as the Wallis formula and is extremely useful for calculating integrals of powers of sine and cosine. The formula works for both $\sin^n x$ and $\cos^n x$ over $[0, \pi/2]$ due to the symmetry proven in Problem 49.