Question
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Answer
Problem 3: 243/5
Problem 4: 0
Problem 5: 0
Problem 4: 0
Problem 5: 0
Solution
Problem 3
We wish to evaluate
Notice that the numerator
is an even function. Although the denominator
is not even, we can use a symmetry trick. Define
and consider
Now, add the two:
Factor out
:
Observe that
so that
Thus,
Since the interval is symmetric, we have
But by a change of variable
the two integrals on the left are equal, so
Now, compute the even function integral:
Thus,
Problem 4
We need to evaluate
Apply the substitution
When
,
; when
,
. The integral becomes
Separate the integral:
Recall the standard integrals:
Adding these, we have
Thus,
At
:
At
:
Hence,
Problem 5
We wish to evaluate
Examine the integrand. The numerator
is an even function since
, and the denominator
is an odd function because
. Therefore, the integrand satisfies
This shows that
is an odd function. An integral of an odd function over a symmetric interval about zero is zero provided the integral is interpreted as a Cauchy principal value:
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Beyond the Answer
To tackle the integrals presented, recognizing symmetry and periodic properties can simplify the calculations significantly!
For the first integral, observe that the integrand
is an even function. This means that
, allowing you to rewrite the integral as
, which can reduce the computational effort.
In the second problem, notice that
and
are periodic functions. You can use substitution to simplify the integration—let
transforming the limits and the function, potentially leading to a straightforward integration over a complete period of sine and cosine functions.
Keeping a keen eye on symmetries or properties of periodic functions can bring clarity, reduce complexity, and help avoid common mistakes like miscalculating the limits or overlooking these traits! Happy integrating!