# Thread: Integrals of Trig Functions

1. ## Integrals of Trig Functions

I'm having trouble with this problem:

tan^2(x^1/2)/x^(1/2)
I don't know how to get any special characters in there so just assume I want to integrate the above problem. Just for clarification...it reads tan squared square root x. (The entire tan function is squared) divided by square root x.

2. Originally Posted by nick898
I'm having trouble with this problem:

I don't know how to get any special characters in there so just assume I want to integrate the above problem. Just for clarification...it reads tan squared square root x. (The entire tan function is squared) divided by square root x.
$\displaystyle \int \frac{tan^2(\sqrt{x})}{\sqrt{x}}~dx$

Let $\displaystyle y = \sqrt{x} \implies dx = 2y~dy$

So
$\displaystyle \int \frac{tan^2(\sqrt{x})}{\sqrt{x}}~dx = \int \frac{tan^2(y)}{y} \cdot 2y~dy$

$\displaystyle = 2 \int tan^2(y)~dy$

How 'bout now?

-Dan

3. Ok. I understand how to get the answer out of that. (It has the answers in the back of the book) but I'm confused how you got dx = -2ydy

I think it's just a minor difference in the way I was taught but I just can't see it.

I understand using a variable to represent square root x. I just don't get the dx = -2ydy part.

4. Originally Posted by nick898
Ok. I understand how to get the answer out of that. (It has the answers in the back of the book) but I'm confused how you got dx = -2ydy

I think it's just a minor difference in the way I was taught but I just can't see it.

I understand using a variable to represent square root x. I just don't get the dx = -2ydy part.
Sorry, I was off by a - sign:
$\displaystyle y = \sqrt{x}$

$\displaystyle dy = \frac{1}{2\sqrt{x}}~dx$

$\displaystyle dy = \frac{1}{2y}~dx$

$\displaystyle dx = 2y~dy$

-Dan

5. Ok, it all makes sense now. Thank you!