# What is the integral of #int (sinx)/(cos^2x) dx#?

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mason m
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May 28, 2018

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Alternatively, you could rewrite this in terms of other trigonometric functions:

#intsinx/cos^2xdx=int(1/cosx)(sinx/cosx)dx=intsecxtanxdx#

If you're familiar with the fact that

#intsecxtanxdx=secx+C#

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mason m
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Mar 29, 2017

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We should try to use substitution by setting

This gives us the integral:

#intsinx/cos^2xdx=-int(-sinx)/cos^2xdx=-int1/u^2du=-intu^-2du#

From here, use the rule

#intu^ndu=u^(n+1)/(n+1)+C#

Thus,

#-intu^-2du=-u^(-1)/(-1)+C=1/u+C#

#=1/cosx+C=secx+C#

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Konstantinos Michailidis
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Mar 17, 2016

It is

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