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

3 Answers

It is

#int (sinx)/(cos^2x) dx= int [1/cosx]'dx=1/cosx+c=secx+c#

Mar 17, 2016

Answer:

#secx+C#

Explanation:

We should try to use substitution by setting #u=cosx#, so #du=-sinxdx#.

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

Answer:

#secx+C#

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 #d/dx(secx)=secxtanx#, then you'll recognize this common integral formula:

#intsecxtanxdx=secx+C#