How do you find the integral of #int [cot^5 (x) (sin^4(x) dx]#?

Answer 1

#I=ln|sinx|-sin^2x+sin^4x/4+C#

#int cot^5xsin^4xdx=int (cos^5x)/(sin^5x)sin^4xdx=#
#int (cos^4xcosx)/(sinx)dx=int ((1-sin^2x)^2cosx)/sinx dx=I#
#sinx=t => cosxdx=dt#
#I=int (1-t^2)^2/tdt=int (1-2t^2+t^4)/tdt=int (1/t-2t+t^3)dt#
#I=ln|t|-t^2+t^4/4+C#
#I=ln|sinx|-sin^2x+sin^4x/4+C#
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Answer 2

To find the integral of ( \int \cot^5(x) \sin^4(x) , dx ), you can use trigonometric identities to simplify the integral. One common approach is to rewrite ( \cot^5(x) ) in terms of ( \sin(x) ) and ( \cos(x) ), and then use substitution to simplify the integral. Another approach involves rewriting ( \sin^4(x) ) using the power-reducing identity and then proceeding with integration. After simplification, you can integrate the resulting expression term by term.

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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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