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How do you find the integral of #(cosx)^2/(sinx)#?

Answer 1

Emily Ammerman

Use #cos^2x/sinx = cscx-sinx #

#cos^2x/sinx = (1-sin^2x)/sinx = 1/sinx - sin^2x/sinx = cscx-sinx #

#int sinx dx = -cosx +C#

#int cscx dx = -ln abs(cscx+cotx) +C# You can either memorize this untegral of the trick to getting it:

#int cscx dx = int cscx (cscx+cotx)/(cscx+cotx) dx#

# = - int (-csc^2x-cscx cotx)/(cscx+cotx)dx#

# = -int 1/u du = -ln abs u +C#

I think you can finish #int cos^2x/sinx dx# from here.

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Answer 2

Isaiah Allen

To find the integral of ( \frac{{\cos^2(x)}}{{\sin(x)}} ), you can use the substitution method. Let ( u = \sin(x) ), then ( du = \cos(x) dx ). Substitute ( u ) and ( du ) into the integral, and rewrite the integral in terms of ( u ). You'll end up with ( \int \frac{{1 - u^2}}{{u}} du ). Divide ( u ) into ( 1 - u^2 ), and then integrate term by term. You will obtain ( \ln|\sin(x)| - \sin(x) + C ), where ( C ) is the constant of integration.

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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.