The binomial theorem states that (a + b)n =n∑k=0nCkan−kbk for all n ∈ N.Use the binomial theorem to help show that∫ π20 cos2n+1 x dx =n∑k=0nCk(−1)k2k + 1
Question
The binomial theorem states that (a + b)n =n∑k=0nCkan−kbk for all n ∈ N.Use the binomial theorem to help show that∫ π20 cos2n+1 x dx =n∑k=0nCk(−1)k2k + 1
Solution
The proof of the given equation involves a few steps. Here's how you can do it:
Step 1: Start with the integral ∫ cos^(2n+1)(x) dx from 0 to π/2.
Step 2: Use the power-reducing identity cos^2(x) = (1 + cos(2x))/2 to rewrite the integral.
Step 3: Expand the integral using the binomial theorem, which states that (a + b)^n = Σ (from k=0 to n) nCk * a^(n-k) * b^k.
Step 4: In this case, a = 1 and b = cos(2x), so the integral becomes ∫ (from 0 to π/2) Σ (from k=0 to n) nCk * cos^(2k)(2x) dx.
Step 5: Now, integrate term by term. The integral of cos^(2k)(2x) from 0 to π/2 is 0 for all k > 0, because the function completes an integer number of cycles over this interval.
Step 6: The only term that does not vanish is the k = 0 term, which gives ∫ (from 0 to π/2) nC0 dx = π/2.
Step 7: Therefore, ∫ (from 0 to π/2) cos^(2n+1)(x) dx = Σ (from k=0 to n) nCk * (−1)^k / (2k + 1) * π/2.
Step 8: Multiply both sides by 2 to get the final result: ∫ (from 0 to π/2) cos^(2n+1)(x) dx = Σ (from k=0 to n) nCk * (−1)^k / (2k + 1).
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a Let k ∈ N. Find an expression in terms of k for ∫ π20 sin2k x dx
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