Wallis's Formulas state that
1) If $n$ is odd and $n \geq 3,$ then
$$ \int _0 ^{\pi/2} \cos^nx \ dx= \frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} … \frac{n-1}{n}$$
2) If $n$ is even and $n \geq 2,$ then
$$ \int _0 ^{\pi/2} \cos^nx \ dx= \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} … \frac{n-1}{n} \cdot \frac{\pi}{2}$$
These formulas are also valid when $\cos^nx \ dx$ is replaced by $\sin^nx \ dx.$
I proved both of the forms, but only in terms of cosine. How can I prove that both of the definite integrals are correct for sine function as well? In other words, I am stuck on how to prove
$$
\int _0 ^{\pi/2} \cos^nx \ dx= \int _0 ^{\pi/2} \sin^nx \ dx
$$
other than the obvious fact that the regions look about the same, just "flipped." (we can "flip" the area of the graph of cosine left to right to make it look like the sine function graph).
Best Answer
Use the fact that $$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$ To get: $$\int_0^{\pi/2}\cos^n x dx=\int_0^{\pi/2}\cos^n\left(\frac{\pi}{2}-x\right) xdx=\int_0^{\pi/2}\sin^n x dx$$