91Ó°ÊÓ

Based on the given step by step solution, answer the following question: **Question:** Prove the integral identity: $$\int_{0}^{\pi / 2} \cos ^{m} x \sin^{m} x d x = 2^{-m} \int_{0}^{\pi / 2} \cos^{m} xdx$$ **Answer:** We can prove the integral identity using the following steps: 1. Rewrite the given integrals in terms of a single variable, resulting in: $$\int_{0}^{\pi / 2} \cos ^{m} x \sin^{m} x d x = 2^{-m} \int_{0}^{\pi / 2} \cos^{m} x dx$$ 2. Apply the substitution \(u = \cos x\) and find its differential: $$du = -\sin x dx$$ 3. Solve for \(\sin x dx\) and plug in the substitution into the integral: $$\int_{1}^{0} u^{m} (-du)$$ 4. Simplify the new integral: $$\int_{0}^{1} u^{m} du$$ 5. Return to the original context and compare the two obtained expressions: $$\int_{0}^{\pi / 2} \cos ^{m} x \sin^{m} x d x = \int_{0}^{1} u^{m} du$$ and $$2^{-m} \int_{0}^{\pi / 2} \cos^{m} xdx = 2^{-m} \int_{0}^{1} u^{m} du$$ 6. The given identity is proven as both expressions are equal: $$\int_{0}^{\pi / 2} \cos ^{m} x \sin^{m} x d x = 2^{-m} \int_{0}^{\pi / 2} \cos^{m} xdx$$