Question

Question #5a465

Answer 1

`int\ sin(x)/cos^2(x)\ dx=1/cos(x)+C=sec(x)+C`

Explanation:

`int\ sin(x)/cos^2(x)\ dx`

Let's substitute `t=cos(x)` and `dt/dx=-sin(x)`.

Then, we have
`int\ sin(x)/cos^2(x)\ dx`
`=int\ sin(x)/cos^2(x)dx/dt\ dt`
`=-int\ sin(x)/t^2 1/sin(x)\ dt`
`=-int\ dt/t^2`
`=1/t+C`

Substitute back `t=cos(x)` to get the final answer of
`=1/cos(x)+C`

Another way to solve the integration problem is to use the fact that `tan(x)=sin(x)/cos(x)` and `sec(x)=1/cos(x)` to get that
`int\ sin(x)/cos^2(x)\ dx`
`=int\ tan(x)sec(x)\ dx`

Recall that `d/dx(sec(x))=tan(x)sec(x)`. So, we have the above integral equal to
`=sec(x)+C`

As seen, both two methods get the same answer.