Question

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

Answer 1

It is

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

Answer 2

`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`

Answer 3

`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`