# Thread: Integral of Trigonometric Functions

1. ## Integral of Trigonometric Functions

Hi, sorry for not fixing my profile but I'll get right on it after posting this problem. I'm sure there is a rule about Latex here but forgive me if I will post the problem through an image. I hope that is okay.

Here is my problem:

I do not know where to start or perhaps I could not wrap around my head in this problem. I tried thinking of ways but I really can't and I am stumped. Our current and past lessons were Inverse Trigonometric Functions and Powers of Sine, Cosine, Tangents (Integrals, of course). I even tried Integration by parts but I still can't start how. So, this is my last resort.

2. $\displaystyle \frac{\cot{\theta}}{\csc{\theta} + 4\sin{\theta}} = \frac{\frac{\cos{\theta}}{\sin{\theta}}}{\frac{1}{ \sin{\theta}} + 4\sin{\theta}}$

$\displaystyle = \frac{\cos{\theta}}{1 + 4\sin^2{\theta}}$

So $\displaystyle \int{\frac{\cot{x}}{\csc{x} + 4\sin{x}}\,dx} = \int{\frac{\cos{x}}{1 + 4\sin^2{x}}\,dx}$.

Now let $\displaystyle u = \sin{x}$ so that $\displaystyle du = \cos{x}\,dx$ and the integral becomes

$\displaystyle \int{\frac{1}{1 + 4u^2}\,du}$.

Now we need to make a trigonometric substitution $\displaystyle u = \frac{1}{2}\tan{\theta}$ so that $\displaystyle du = \frac{1}{2}\sec^2{\theta}\,d\theta$ and $\displaystyle \theta = \arctan{2u}$ and the integral becomes

$\displaystyle \int{\frac{1}{1 + 4\left(\frac{1}{2}\tan{\theta}\right)^2}\left(\fra c{1}{2}\sec^2{\theta}\right)\,d\theta}$

$\displaystyle = \frac{1}{2}\int{\frac{\sec^2{\theta}}{1 + \tan^2{\theta}}\,d\theta}$

$\displaystyle = \frac{1}{2}\int{\frac{\sec^2{\theta}}{\sec^2{\thet a}}\,d\theta}$

$\displaystyle = \frac{1}{2}\int{1\,d\theta}$

$\displaystyle = \frac{1}{2}\theta + C$

$\displaystyle = \frac{1}{2}\arctan{2u} + C$

$\displaystyle = \frac{1}{2}\arctan{(2\sin{x})} + C$.

Where did you get $\displaystyle u = \frac{1}{2}\tan{\theta}$ from if $\displaystyle u = \sin{x}$?

4. It's just a renaming...

5. I see, I'm just not accustomed to that kind of substitution. Trigonometric substitution was taught in a different way in what I saw in your post.

If $\displaystyle a^2+u^2$ in substitution, shouldn't it be $\displaystyle u = (a)tan{\theta}$

or you can substitute anything?

I got the first part by the way and thanks for that, the second part is a little bit hazy and I really want to learn more.

6. If you wanted to do it that way, you would have to take out a factor of $\displaystyle \frac{1}{4}$ to get it of that form.

Of course then, you would have an extra factor to take out after you've made the substitution. The way I posted avoids taking out 2 extra common factors...

7. If I have followed my formula, I would've ended with this:

$\displaystyle \int{\frac{\sec^2{\theta}}{1 + (4)\tan^2{\theta}}\,d\theta}$

I really appreciate your shortcut but I really need the longer method if that's alright. I really want to understand this better in order for me to teach it to my classmates as well.

Where do I go from here?
$\displaystyle \int{\frac{\sec^2{\theta}}{1 + (4)\tan^2{\theta}}\,d\theta}$

8. No, that's not correct. Like I said, if you want to use the longer method you need to take out the common factor of $\displaystyle \frac{1}{4}$ in the denominator, because you need it to be of the form $\displaystyle a^2+ u^2$, not $\displaystyle a^2 + 4u^2$...

9. I'm going to sound a total newbie but... is this correct?

EQUATION ERASED. WRONG EQUATION.

---------------------
Oh crap, I just forgot factoring. =_=

Still sorting it out.

10. $\displaystyle 1+4u^2$

Factoring this, I will have

$\displaystyle 4 (\frac{1}{2}^2+u^2)$

So basically, that's where you got:
$\displaystyle u = \frac{1}{2}\tan{\theta}$

Am I right?

11. Correct.

And then after you have simplified this, you will need to factorise again to get $\displaystyle 1 + \tan^2{\theta}$ which simplifies to $\displaystyle \sec^2{\theta}$.

12. Thank you very much! Thanks for being patient with me.