Question

How do you find the integral of `((1+sinx)/cosx)dx`?

Answer 1

`ln|tanx+secx|-ln|cosx|+C`

Explanation:

We can start by rewriting our integral as

`int (1/cosx)+color(blue)((sinx/cosx)) dx`

What I have in blue is equal to `tanx`, so now we have the following integral:

`int (1/cosx+tanx) dx`

The integral of the sum is equal to the sum of the integrals, so we can rewrite the above expression as

`color(springgreen)(int (1/cosx)dx)+color(orchid)(int (tanx)dx)`

Integrating `1/cosx`

`1/cosx` is equivalent to `secx`. Now, we have the integral

`color(springgreen)(int (secx) dx)`, which evaluates to

`color(springgreen)(ln|tanx+secx| +C)`

Remember, this is one part of our integral.

Integrating `tanx`

We can rewrite `tanx` as `sinx/cosx`. Doing this, we get

`color(orchid)(int (sinx/cosx) dx)`

Whenever we see a function and its derivative, it's a good idea to use u-substitution. Let's define

`color(orchid)(u=cosx)`

From this, we know that

`color(orchid)(du=-sinx)`

Since we have `sinx`, not `-sinx`, we can put a negative in front of the integral. Plugging in, we get

`color(orchid)(-int (1/u du))`

which evaluates to

`color(orchid)(-ln|u|)`

Now, we need to back-substitute (plug `u` back in). We get

`color(orchid)(-ln|cosx|+C)`

Thus, the integral of `((1+sinx)/cosx) dx` is

`bar (ul( |color(white)(2/2)(ln|tanx+secx|-ln|cosx|+C)color(white)(2/2) |`

Hope this helps!