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How do you find #int sec^2x/(1-sinx) #?

Answer 1

Charles Arocha

=#tanx +tan^3x/3+sec^3x/3+C#

#int(sec^2x)/(1-sinx)dx# #=int(sec^2x)/(1-sinx)*(1+sinx)/(1+sinx)dx#

#=int(sec^2x)/(1-sin^2x)*(1+sinx)dx#

#=int(sec^2x/cos^2x)*(1+sinx)dx#

#=intsec^2x*(1/cos^2x+sinx/cos^2x)dx# #=intsec^2x*sec^2xdx+intsec^2xsecxtanxdx#

#=intsec^2x*(1+tan^2x)dx+intsec^2xsecxtanxdx# For first part let #u=tanx =>du=sec^2xdx# For second part let #v = secx =>dv=secxtanxdx#

The whole integral becomes =#int(1+u^2)du+intv^2dv# =#u +u^3/3+v^3/3+C#

=#tanx +tan^3x/3+sec^3x/3+C#

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Answer 2

Christopher Agresta

To find the integral of sec^2(x) / (1 - sin(x)) with respect to x, you can use trigonometric identities and substitution. Let u = 1 - sin(x), then du = -cos(x)dx.

The integral becomes ∫(sec^2(x) / (1 - sin(x))) dx = -∫(1/u) du = -ln|u| + C, where C is the constant of integration.

Substituting back for u, we get -ln|1 - sin(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.