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How do you find the indefinite integral of #int (cot(x))^(1/26) csc(x)^2 dx#?

Answer 1

Emma Aldridge

#I=-26/27cotxroot(26)(cotx) +C#

I presume that you mean #(cscx)^2#

#cotx=t => -csc^2xdx=dt#

#I=int (cotx)^(1/26)(cscx)^2dx=-int t^(1/26)dt=-t^(27/26)/(27/26)+C#

#I=-26/27cotxroot(26)(cotx) +C#

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Answer 2

Luke Alvarez

To find the indefinite integral of ( \int (\cot(x))^{1/26} \csc^2(x) , dx ), we can use trigonometric identities and substitution.

Let ( u = \cot(x) ), then ( du = -\csc^2(x) , dx ).

Now, rewrite the integral in terms of ( u ):

[ \int u^{1/26} , du ]

Apply the power rule for integration:

[ \frac{u^{1/26 + 1}}{1/26 + 1} + C ]

[ = \frac{26}{27} u^{27/26} + C ]

Finally, substitute back ( u = \cot(x) ):

[ = \frac{26}{27} (\cot(x))^{27/26} + C ]

So, the indefinite integral of ( \int (\cot(x))^{1/26} \csc^2(x) , dx ) is ( \frac{26}{27} (\cot(x))^{27/26} + C ).

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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.