Prove the following equality: integral from 0 to (pi/2)

Prove the following equality: $\int_0^{\pi/2} f(\sin x) \, dx = \int_0^{\pi/2} f(\cos x) \, dx$.

Proof using substitution:

Let $u = \frac{\pi}{2} - x$. Then $du = -dx$, so $dx = -du$; also $x = \frac{\pi}{2}-u$.

Change the limits of integration:

When $x = 0$: $u = \frac{\pi}{2}$. When $x = \frac{\pi}{2}$: $u = 0$.

Express trigonometric functions in terms of $u$:

Using the co-function identity:

\[\sin x = \sin\left(\frac{\pi}{2} - u\right) = \cos u\]

Transform the integral:

\[\int_0^{\pi/2} f(\sin x) \, dx = \int_{\pi/2}^0 f(\cos u) \cdot (-du) = \int_0^{\pi/2} f(\cos u) \, du\]

Since the variable of integration is a dummy variable, we can replace $u$ with $x$:

\[= \int_0^{\pi/2} f(\cos x) \, dx\]

Therefore:

\[\boxed{\int_0^{\pi/2} f(\sin x) \, dx = \int_0^{\pi/2} f(\cos x) \, dx}\]

Geometric interpretation:

This equality reflects the complementary relationship between sine and cosine. On $[0, \frac{\pi}{2}]$: as $x$ goes from $0$ to $\frac{\pi}{2}$, $\sin x$ increases from $0$ to $1$ while $\cos x$ decreases from $1$ to $0$; the transformation $u = \frac{\pi}{2} - x$ swaps sine and cosine, and both functions traverse the same set of values $[0, 1]$, just in opposite order.

Example applications:

Example 1: Show that $\int_0^{\pi/2} \sin^2 x \, dx = \int_0^{\pi/2} \cos^2 x \, dx$. Using our property with $f(t) = t^2$, and since $\sin^2 x + \cos^2 x = 1$:

\[2\int_0^{\pi/2} \sin^2 x \, dx = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \quad\Rightarrow\quad \int_0^{\pi/2} \sin^2 x \, dx = \frac{\pi}{4}\]

Example 2: Evaluate $I = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx$. Using our property, $I = \int_0^{\pi/2} \frac{\cos x}{\sin x + \cos x} \, dx$. Adding these two expressions:

\[2I = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \quad\Rightarrow\quad I = \frac{\pi}{4}\]

Example 3: For any positive integer $n$: $\int_0^{\pi/2} \sin^n x \, dx = \int_0^{\pi/2} \cos^n x \, dx$.

Connection to Problem 44:

This result is actually a special case of Problem 44 applied to the interval $[0, \frac{\pi}{2}]$: with $a = 0$ and $b = \frac{\pi}{2}$, we have $a + b - x = \frac{\pi}{2} - x$, and $\sin\left(\frac{\pi}{2} - x\right) = \cos x$, so

\[\int_0^{\pi/2} f\left(\sin\left(\frac{\pi}{2} - x\right)\right) dx = \int_0^{\pi/2} f(\sin x) \, dx\]

which becomes $\int_0^{\pi/2} f(\cos x) \, dx = \int_0^{\pi/2} f(\sin x) \, dx$.