# Thread: [SOLVED] Simple integration problem

1. ## [SOLVED] Simple integration problem

$\displaystyle \int sec(3x) tan(3x) dx$

It's an easy problem, yet... I don't know how to go about solving it. Can anyone shed some light?

2. Here is a hint: $\displaystyle \sec x\tan x$ is the derivative of something.

Hope this helps!

3. Right, sec 3x tan 3x becomes sec 3x.

What confuses me is what's to be done after that.

=

$\displaystyle \int sec(3x)dx$

Do I use the substitution rule?

4. Let $\displaystyle u=3x\mbox{ Then }du=3dx$. Then upon substitution you get.

$\displaystyle \frac{1}{3}\int sec(u)tan(u)\,du$

Go from there.

5. How can I proceed? Do I use the general power rule?

I end up with sec $\displaystyle 3x^2/6 + C$ which apparently is not the answer.

6. go through your inventory of the derivatives of the trig functions. such as $\displaystyle \frac{d}{dx}sec(x)$

7. Could you please be a little bit more specific? is $\displaystyle sec x tan x$ ... how does that help me? I already did that step. My final answer is wrong though.

8. If the derivative of $\displaystyle sec(x)$ is $\displaystyle sec(x)tan(x)$, then what is the integral of $\displaystyle sec(x)tan(x)$?

9. sec x.

But I already did that part. Why are you bringing it up?

10. by substitutin u in you get sec(u) as there is 1/3 outside in the integral the final answer comes to 1/3*sec(3x)

11. well, you have $\displaystyle \frac{1}{3}\int sec(u)tan(u)\,du$, so after you integrate that, and substitute $\displaystyle 3x$ back in for $\displaystyle u$, what do you think?

12. Originally Posted by Fermatwannabe
by substitutin u in you get sec(u) as there is 1/3 outside in the integral the final answer comes to 1/3*sec(3x)
Thank you

13. Well that is the answer to the original integral you posted. Can you double check it? Unless your answer is $\displaystyle \frac{1}{cos(3x)}+C$

14. Originally Posted by Archduke01
Thank you
is it 1/cos(3x) if not then im stumped

15. Originally Posted by Archduke01
$\displaystyle \int sec(3x) tan(3x) dx$

It's an easy problem, yet... I don't know how to go about solving it. Can anyone shed some light?
You are trying to find the antiderivative of $\displaystyle \sec{3x}\tan{3x}$

The derivative of $\displaystyle sec{3x}$ is $\displaystyle 3\sec{3x}\tan{3x}$

So the antiderivative of $\displaystyle 3\sec{3x}\tan{3x}$ is $\displaystyle sec{3x}+C$

Then what would be the antiderivative of $\displaystyle \sec{3x}\tan{3x}$?