Question

How do you find the indefinite integral of `int tan5theta`?

Answer 1

`int tan5theta d theta = 1/5 ln|sec5 theta| + C`

Explanation:

Let `u=5 theta => (du)/(d theta)=5`
So, `int ...du = int ... 5d theta`, or `int ... d theta = int ... 1/5du`

Substituting into `int tan5theta d theta` we get:

`int tan5theta d theta = int tanu 1/5 du`
`:. int tan5theta d theta = 1/5 int tanu du`
`:. int tan5theta d theta = 1/5 ln|secu| + C`

And then replacing `u=5 theta` gives:

`int tan5theta d theta = 1/5 ln|sec5 theta| + C`

Answer 2

`=-1/5 lncos 5theta +C`, in `Q_1 and Q_3` and

`=-1/5ln(-cos 5theta)`, in `Q_2 and Q_4`..

Explanation:

`tan 5theta` is continuous for any interval sans (not including )

`5theta`=an odd multiple of `pi/2`.

So, in any such interval.

the indefinite `int tan 5theta d theta`

`=int d(-1/5ln cos 5theta)`, in `Q_1 and Q_3`

`=-1/5 lncos 5theta +C`, in `Q_1 and Q_3` and

`=int d(-1/5ln(- cos 5theta)`, in `Q_2 and Q_4`

`=-1/5ln(-cos 5theta)`, in `Q_2 and Q_4`..

Note that as `theta to( (2k-1)pi/10)_(+-),`

`5theta to((2k-1)pi/2)_(+-) and tan 5theta to +-oo`