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

Prove: 1) $\int_0^{\pi/2} \sin^n x \, dx = \int_0^{\pi/2} \cos^n x \, dx$. 2) $\int_0^{\pi/2} \frac{\sqrt{\sin^{2k} x + \cos^{2k} x}}{\cos 2x} \, dx = 0$.

Part $1^{\circ}$: Prove $\int_0^{\pi/2} \sin^n x \, dx = \int_0^{\pi/2} \cos^n x \, dx$

Proof using substitution:

Let $u = \frac{\pi}{2} - x$, so $du=-dx$, $dx=-du$, $x=\frac\pi2-u$.

Change the limits of integration:

When $x=0$: $u=\frac\pi2$. When $x=\frac\pi2$: $u=0$.

Use the co-function identity: $\sin x = \sin\left(\frac{\pi}{2} - u\right) = \cos u$.

Transform the integral:

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

Therefore:

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

Note: This result is the special case of Problem 45 with $f(t) = t^n$.

Part $2^{\circ}$: Prove $\int_0^{\pi/2} \frac{\sqrt{\sin^{2k} x + \cos^{2k} x}}{\cos 2x} \, dx = 0$

Proof:

Let $I = \int_0^{\pi/2} \frac{\sqrt{\sin^{2k} x + \cos^{2k} x}}{\cos 2x} \, dx$. We use the substitution $u = \frac{\pi}{2} - x$ (as in Part 1).

Step 1: Apply the substitution

When $x=0$: $u=\frac\pi2$. When $x=\frac\pi2$: $u=0$. And $dx=-du$.

Step 2: Transform the integrand

For the numerator: $\sin x = \cos u$ and $\cos x = \sin u$, so

\[\sin^{2k} x + \cos^{2k} x = \cos^{2k} u + \sin^{2k} u\]

For the denominator:

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

Step 3: Rewrite the integral

\[I = \int_{\pi/2}^0 \frac{\sqrt{\sin^{2k} u + \cos^{2k} u}}{-\cos(2u)} \cdot (-du) = \int_{\pi/2}^0 \frac{\sqrt{\sin^{2k} u + \cos^{2k} u}}{\cos(2u)} \, du\]

\[= -\int_0^{\pi/2} \frac{\sqrt{\sin^{2k} u + \cos^{2k} u}}{\cos(2u)} \, du\]

Since $u$ is a dummy variable:

\[I = -\int_0^{\pi/2} \frac{\sqrt{\sin^{2k} x + \cos^{2k} x}}{\cos(2x)} \, dx = -I\]

Step 4: Solve for $I$

\[I = -I \quad\Rightarrow\quad 2I = 0 \quad\Rightarrow\quad I = 0\]

Therefore:

\[\boxed{\int_0^{\pi/2} \frac{\sqrt{\sin^{2k} x + \cos^{2k} x}}{\cos 2x} \, dx = 0}\]

Geometric interpretation:

The result $I=-I$ means the integral has odd symmetry under $x \to \frac{\pi}{2} - x$: the numerator $\sqrt{\sin^{2k}x+\cos^{2k}x}$ is symmetric, but the denominator $\cos 2x$ flips sign, so the integrand changes sign under this transformation and the positive/negative contributions cancel exactly.

Alternative view:

Splitting at $x=\frac\pi4$: on $[0,\frac\pi4)$, $\cos 2x>0$ (since $2x\in[0,\frac\pi2)$); on $(\frac\pi4,\frac\pi2]$, $\cos 2x<0$ (since $2x\in(\frac\pi2,\pi]$). The symmetry property shows these two parts cancel each other.