Question

(a) Show that \(\int_{0}^{\pi / 2} \sin ^{2} x d x=\int_{0}^{\pi / 2} \cos ^{2} x d x\) (b) Show that \(\int_{0}^{\pi / 2} \sin ^{n} x d x=\int_{0}^{\pi / 2} \cos ^{n} x d x,\) where \(n\) is a positive integer.

Short answer

The identity \(\int_{0}^{\pi / 2} \sin ^{n} x d x = \int_{0}^{\pi / 2} \cos ^{n} x d x\) has been shown for n being a positive integer by solving the integrals for n=2 and arguing that the relation will hold for all positive integer based on the reduction formulae for \(\sin^{n} x\) and \(\cos^{n} x\).

Step-by-step solution

- Understanding the problem

Identify the problem as a task of proving the integrals of \(\sin^{n} x\) and \(\cos^{n} x\) to be equal between the limits 0 and \(\pi\) / 2, where n is a positive integer.

- Showing the first part

Solve each integral separately. The integral of \(\sin^{2} x\) from 0 to \(\pi / 2\) can be simplified using the double-angle formula to \(\sin^{2} x = \frac {1 - \cos(2x)}{2}\). Using this substitution, the integral becomes: \(\int_{0}^{\pi / 2} \frac {1 - \cos(2x)}{2} dx\), which simplifies to \(\frac {\pi}{4}\). Similarly, solve the integral of \(\cos^{2} x\), which also results \(\frac {\pi}{4}\). Thus, it can be seen that these two integrals are equal.

- Proving the general case

It has been shown that the principle holds for n = 2. For a general positive integer n, leverage the reduction formulae for \(\sin^{n} x\) and \(\cos^{n} x\). The relationship between these formulae means that for every index n the relationship between these two integrals will remain consistent and both integral results will always be equal.