# Math Help - Integration

1. ## Integration

1. Integrate following with respect to x:
sech 2x

2. Find the area of the finite region bounded by the curve with equation x^2 – y^2 = 4 and the line x = 5.

2. $\int{sech(2x)}dx$

Rewrite $sech(2x)=\frac{2e^{2x}}{e^{4x}+1}$

Let $u=e^{2x}, \;\ \frac{du}{2}=e^{2x}dx$

$\int\frac{1}{u^{2}+1}du$

Now, see what the antiderivative is?.

#2: Solve your equation for x in terms of y and integrate from 0 to 5.

$2\int_{0}^{5}\sqrt{4+y^{2}}dy$

3. Originally Posted by galactus
[tex]

#2: Solve your equation for x in terms of y and integrate from 0 to 5.

$2\int_{0}^{5}\sqrt{4+y^{2}}dy$
There is a problem, I can’t solve this.

Ok, here is my work,

Let, I = ∫(4+y^2)^1/2
Let, x = (4+y^2)^3/2
dx/dy = 3y (4+y^2)^1/2

So: I = 1/3y (4+y^2)^3/2 + C

With limits 0 – 5, I got the answer 10.41. But correct answer is 16.6.

4. Originally Posted by geton
There is a problem, I can’t solve this.

Ok, here is my work,

Let, I = ∫(4+y^2)^1/2
Let, x = (4+y^2)^3/2
dx/dy = 3y (4+y^2)^1/2

So: I = 1/3y (4+y^2)^3/2 + C

With limits 0 – 5, I got the answer 10.41. But correct answer is 16.6.
where did you learn that integration technique?

use a trig substitution, $y = 2 \tan \theta$

5. Originally Posted by Jhevon
where did you learn that integration technique?
I did a mistake.

use a trig substitution, $y = 2 \tan \theta$
By the way, trig substitution, can you explain for me? I don’t get yet.

6. Originally Posted by geton
By the way, trig substitution, can you explain for me? I don’t get yet.
do you mean you haven't done it in class, or you're just not good at it. in either case, look it up in your textbook. once you figure out what to substitute, it's pretty much the regular substitution technique. back-substituting can be somewhat annoying, but it is not that bad, and it can be avoided altogether with definite integrals as you see here.

so let's start you off:

we wish to evaluate $\int_0^5 \sqrt{4 + y^2}~dy$

Let $y = 2 \tan \theta$

$\Rightarrow dy = 2 \sec^2 \theta ~d \theta$

Change our limits to avoid back-substitution:

when $y = 0$, $\theta = \arctan \frac 02 = 0$

when $y = 5$, $\theta = \arctan \frac 52$

So our integral becomes:

$4 \int_0^{\arctan 5/2} \sec^3 \theta ~d \theta$

Now continue