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

1 Answer
Jun 6, 2018

#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!