What is the integral of #int (sinx)/(cos^2x) dx#?
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"Suppose that I don't have a formula for #g(x)# but I know that #g(1)
= 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"
3 Answers
Mar 13, 2016
It is
Mar 17, 2016
Explanation:
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#
Mar 17, 2016
Explanation:
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#
