What is #f(x) = int xcotx^2 dx# if #f((5pi)/4) = 0 #?
I have tried , but it is complicated for
Here,
So,
Given that,
Now,
#(25xxpi^2)/16 ~~15.42....#
Thus,
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To find ( f(x) = \int x \cot(x^2) , dx ) when ( f \left( \frac{5\pi}{4} \right) = 0 ), we first differentiate ( f(x) ) using the fundamental theorem of calculus and apply the given condition to solve for the constant of integration.
Differentiating ( f(x) ), we get:
[ f'(x) = \cot(x^2) - 2x^2 \csc^2(x^2) ]
Given ( f \left( \frac{5\pi}{4} \right) = 0 ), we substitute ( x = \frac{5\pi}{4} ) into ( f(x) ):
[ \int_{0}^{\frac{5\pi}{4}} x \cot(x^2) , dx = 0 ]
[ \left[ -\frac{1}{2} \csc(x^2) \right]_0^{\frac{5\pi}{4}} = 0 ]
[ -\frac{1}{2} \csc \left(\frac{25\pi^2}{16}\right) + \frac{1}{2} \csc(0) = 0 ]
[ -\frac{1}{2} \csc \left(\frac{25\pi^2}{16}\right) = -\frac{1}{2} \csc(0) ]
[ \csc \left(\frac{25\pi^2}{16}\right) = \csc(0) ]
Since ( \csc(0) ) is undefined, there might be an error in the given problem or the function ( f(x) ) might not satisfy the conditions stated.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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