If ∫−2π2π2(x+e)225x2cos2xdx=π(απ2+β), α,β∈Z, then (α+β)2 equals
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Step-by-Step Solution
Step 1: Analyze the integrand and limits
The given integral has symmetric limits from −2π to 2π. This suggests using properties of definite integrals for even and odd functions. However, the integrand is not a simple even or odd function due to the (x+e)2 term.
Step 2: Apply the property ∫abf(x)dx=∫abf(a+b−x)dx
Using the property ∫abf(x)dx=∫abf(a+b−x)dx, where a=−2π and b=2π, we have a+b−x=−x. Substituting this into the integrand, we get a new expression for I. Since cos(−x)=cosx and (−x)2=x2, the numerator remains 25x2cos2x.
Step 3: Combine the original and transformed integrals
Let the original integral be I1 and the transformed integral be I2. Since I1=I2=I, we can add them: 2I=I1+I2. This allows us to combine the denominators. We factor out the common term 25x2cos2x.
Step 4: Simplify the combined fraction
We simplify the sum of the fractions by finding a common denominator. Expanding the terms in the numerator and denominator, we notice that the 2xe terms cancel out, leading to a simplified expression. This simplification is crucial for further steps.
Step 5: Re-evaluate the problem statement
Upon simplifying the sum of the two integrals, we arrive at an integrand that is still complex and does not appear to simplify further into a form that would yield the expected result π(απ2+β) with integer α and β. This suggests that the problem statement might contain a typo, possibly in the denominator or the constant factor, as the current form is not standard for competitive exams or typical integral properties.