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How do you find the integral of #int (sin x)/(cos^2x + 1)dx#?

Answer 1

Charles Anctil

#I = -arctan (cosx) + C#

#I = int sinx/(cos^2x+1)dx = int (sinxdx)/(cos^2x+1) = -int (d(cosx))/(cos^2x+1)#

#I = -arctan (cosx) + C#

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Answer 2

Ezra Adams

To find the integral of ( \int \frac{\sin x}{\cos^2 x + 1} , dx ), you can use a substitution method. Let ( u = \cos x ). Then ( du = -\sin x , dx ). Substituting these into the integral, we get:

[ \int \frac{-1}{u^2 + 1} , du ]

This integral is recognizable as the arctangent function. Therefore, the integral becomes:

[ -\arctan(u) + C ]

Substituting back ( u = \cos x ), we get:

[ -\arctan(\cos x) + C ]

So, the integral of ( \int \frac{\sin x}{\cos^2 x + 1} , dx ) is ( -\arctan(\cos 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.