Question

How do you evaluate the integral `int tan(3x-5)dx`?

Answer 1

`1/3ln|sec(3x-5)|+C.`

Explanation:

If we subst. `(3x-5)=t," then, "3dx=dt.`

`:. I=inttan(3x-5)dx=1/3inttan(3x-5)(3)dx,`

`=1/3inttan t dt,`

`=1/3ln|sec t|,`

`rArr I=1/3ln|sec(3x-5)|+C.`

Answer 2

`\frac{ln|sec(3x - 5)|}{3} + C`

Explanation:

We're going to use u-substitution:

`u = 3x - 5`
`du = 3 dx`

Keep in mind the above statement also means that:

`\frac{1}{3}du = dx`

After u-substitution the integral becomes:

`\int \frac{1}{3}tan(u) du`

We can take out the `\frac{1}{3}`:

`\frac{1}{3} \int tan(u) du`

Which we know is equal to:

`\frac{1}{3}ln|sec(u)| + C`

Now you just plug in the original value of u:

`\frac{1}{3}ln|sec(3x - 5)| + C`

and that's your answer :P

Answer 3

The answer is `=-1/3ln|cos(3x-5)|+C`

Explanation:

We perform this integral by substitution

Let `u=3x-5`, `=>`, `du=3dx`

`inttan(3x-5)dx=1/3inttanudu`

`=1/3int(sinudu)/cosu`

Let `v=cosu`, `=>`, `dv=-sinudu`

`1/3int(sinudu)/cosu=-1/3int(dv)/v`

`=-1/3ln|v|`

`=-1/3ln|cosu|`

`=-1/3ln|cos(3x-5)|+C`