Home > Calculus > Special Limits Involving sin(x), x, and tan(x)

What is #int sin x/ ( 1-cos^2x)dx#?

Answer 1

Luke Alvarez

#ln|cosecx-cotx|+C#

From trig identities, #sin^2x+cos^2x=1#

Hence the integral may be written and solved as follows :

#intsinx/(1-cos^2x)dx=intsinx/sin^2xdx=int1/sinxdx=intcosecxdx#

#=ln|cosecx-cotx|+C#

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Answer 2

Christopher Allers

To evaluate the integral ( \int \frac{\sin(x)}{1 - \cos^2(x)} , dx ), we can use a trigonometric substitution. Let ( u = \cos(x) ). Then ( du = -\sin(x) , dx ).

Substitute ( u = \cos(x) ) and ( du = -\sin(x) , dx ) into the integral:

[ \int \frac{\sin(x)}{1 - \cos^2(x)} , dx = \int \frac{-1}{1 - u^2} , du ]

This integral can be evaluated using a standard integral formula. The integral of ( \frac{-1}{1 - u^2} ) with respect to ( u ) is ( -\arctan(u) + C ).

Finally, substitute ( u = \cos(x) ) back into the expression:

[ -\arctan(\cos(x)) + 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.