What is #f(x) = int xcotx^2 dx# if #f((5pi)/4) = 0 #?

Answer 1

I have tried , but it is complicated for #x^2=((5pi)/4)^2.#

#f(x)=1/2ln|sinx^2|+c,where, c~~0.6623#

Here,

#f(x)=I=intxcotx^2dx=intcotx^2*xdx#
Let, #x^2=u=>2xdx=du=>xdx=1/2du#

So,

#I=intcotu*1/2du#
#=1/2ln|sinu|+c,where,u=x^2#
#=>f(x)=1/2ln|sinx^2|+c...to(A)#

Given that,

#f((5pi)/4)=0#
#=>1/2ln|sin((5pi)/4)^2|+c=0#
#=>2c=-ln|sin((25pi^2)/16)|#
#=>2c=ln|(1/(sin((25pi^2)/16)))|#
#=>c=1/2ln|(1/(sin((25pi^2)/16)))|...to(1)#

Now,

#(25xxpi^2)/16 ~~15.42....#

#sin((25pi^2)/16)~~0.2659... in[-1,1]#
#1/(sin((25pi^2)/16))~~3.7606...#
#ln|1/(sin((25pi^2)/16))| ~~1.3245...#
#1/2*ln|1/(sin((25pi^2)/16))| ~~0.6623#
Hence, from #(1)#
#c ~~ #0.6623

Thus,

#f(x)=1/2ln|sinx^2|+c,where, c~~0.6623#
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Answer 2

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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Answer from HIX Tutor

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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