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How do I integrate the function #f(x)=cos(x)/(sin^2(x))+sin(x)#?

Answer 1

Emilia Aguilar

Rewrite #cosx/sin^2x# and then integrate term by term.

#cosx/sin^2x=1/sinx cosx/sinx=cscxcotx# So,

#int (cosx/sin^2x+sinx)dx=intcscxcotx dx+int sinx dx#

#=-cscx-cosx+C#.

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Answer 2

Cameron Alexander

Try this:

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Answer 3

Ezra Adams

To integrate the function ( f(x) = \frac{\cos(x)}{\sin^2(x)} + \sin(x) ), you can use the method of substitution. Let ( u = \sin(x) ), then ( du = \cos(x)dx ). After substitution, the integral becomes ( \int \frac{1}{u^2} du + \int u du ).

Integrating each term separately, you get ( -\frac{1}{u} + \frac{u^2}{2} + C ), where ( C ) is the constant of integration.

Now, substituting back ( u = \sin(x) ), you get ( -\frac{1}{\sin(x)} + \frac{\sin^2(x)}{2} + C ).

So, the indefinite integral of ( f(x) ) is ( -\csc(x) + \frac{\sin^2(x)}{2} + 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.