Question

Prove that \(\int_{0}^{\pi / 2} \cos ^{\mathrm{m}} \mathrm{x} \sin ^{\mathrm{m}} \mathrm{xd} \mathrm{x}=2^{-\mathrm{m}} \int_{0}^{\pi / 2} \cos ^{\mathrm{m}} \mathrm{xdx} .\)

Short answer

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: $$\int_{0}^{\pi / 2} \cos ^{m} x \sin^{m} x d x = 2^{-m} \int_{0}^{\pi / 2} \cos^{m} xdx$$

Step-by-step solution

Identify the given integrals and rewrite as a single integral

In order to prove the given identity, we must first rewrite the integrals in terms of a single variable. In this case, we can rewrite the given integral as: $$ \int_{0}^{\pi / 2} \cos ^{m} x \sin^{m} x d x = 2^{-m} \int_{0}^{\pi / 2} \cos^{m} x dx $$

Apply the substitution \(u = \cos x\)

Now, we will apply the substitution \(u = \cos x\), and compute its differential: $$ du = -\sin x dx $$ Next, we solve for \(\sin x dx\): $$ \sin x dx = -du $$ Now we can plug in the substitution into the integral: $$ \int_{0}^{\pi / 2} \cos ^{m} x \sin^{m} x d x = \int_{1}^{0} u^{m} (-du) $$ Notice the limits change due to the substitution \(u = \cos x\).

Simplify the new integral

Now, simplify the new integral and combine the constant factor: $$ \int_{1}^{0} u^{m} (-du) = - \int_{1}^{0} u^{m} du = \int_{0}^{1} u^{m} du $$

Return to the original context and finish

To finish the proof, we will go back to the original context and use the substitution again, this time in the form of \(\cos x = u\): $$ 2^{-m} \int_{0}^{\pi / 2} \cos^{m} xdx = 2^{-m} \int_{0}^{1} u^{m} du $$ Now, compare the two integrals obtained: $$ \int_{0}^{\pi / 2} \cos ^{m} x \sin^{m} x d x = \int_{0}^{1} u^{m} du $$ and $$ 2^{-m} \int_{0}^{\pi / 2} \cos^{m} xdx = 2^{-m} \int_{0}^{1} u^{m} du $$ Since both expressions are equal, we have proven the given identity: $$ \int_{0}^{\pi / 2} \cos ^{m} x \sin^{m} x d x = 2^{-m} \int_{0}^{\pi / 2} \cos^{m} xdx $$