# Integrals

• Dec 2nd 2009, 01:42 AM
funnytim
Integrals
Hello!

I'm having some problems with Calculus. I can't even figure out how to get started, let alone do the problem :S

2)Calculate area enclosed by y^2 = 2x+6 and y = x -1

3) Calculate the integral
$\displaystyle \int \frac {\sin2x}{1+cos^2x} dx$

Thanks guys ;)
• Dec 2nd 2009, 01:52 AM
tonio
Quote:

Originally Posted by funnytim
Hello!

I'm having some problems with Calculus. I can't even figure out how to get started, let alone do the problem :S

2)Calculate area enclosed by y^2 = 2x+6 and y = x -1

3) Calculate the integral
$\displaystyle \int \frac {\sin2x}{1+cos^2x} dx$

Thanks guys ;)

I'll give you a hand with the first one, you try to make some work of your own with the other ones:

Put $\displaystyle F(x):=\int\limits_{\tan x}^{x^2}\frac{1}{\sqrt{2+t^4}}dt$ As the function in the integral is continuous and defined everywhere, the Fundamental Theorem of integral Calculus tells us that

$\displaystyle F(x)=G(x^2)-G(\tan x)$ , where $\displaystyle G$ is a primitive function of the integrand function, but then:

$\displaystyle F'(x)=2xG'(x^2)-\frac{1}{\cos^2x}G'(\tan x)$ $\displaystyle =2x\,\frac{1}{\sqrt{2+x^8}}-\frac{1}{\cos^2x}\,\frac{1}{\sqrt{2+\tan^4x}}$

Tonio
• Dec 2nd 2009, 01:56 AM
simplependulum
For question three .

Consider $\displaystyle \sin(2x) = 2\sin(x) \cos(x)$

Then make a suitable substitution (Happy)
• Dec 2nd 2009, 02:05 AM
Calculus26
1.For your first question see attachment

2. For your second question see second attachment

3.For your third question write sin(2x) = 2sin(x)cos(x)

Then make the substitution u = cos(x) you will obtain -2u/(1+u^2)

in the integrand which you should be able to do
• Dec 3rd 2009, 01:31 AM
funnytim
Thanks guys, you're great!

One question in#2 though. After i obtain:

How do I calculate it to obtain the answer, 18?

Thank you again!
• Dec 3rd 2009, 01:58 AM
Calculus26
Integrate and evaluate at the limits---submit your work if you don't get 18 and I'll see what happened
• Dec 3rd 2009, 11:28 AM
funnytim
Actually, while I'm puzzling over that problem, here's another: How do I find a definite integral?

Thanks again.
• Dec 3rd 2009, 03:51 PM
tonio
Quote:

Originally Posted by funnytim
Actually, while I'm puzzling over that problem, here's another: How do I find a definite integral?

Doesn't it look suspiciously similar to a Riemann sum of the function $\displaystyle \frac{x}{x^2+1}$ over the interval $\displaystyle [0,1]$?(Giggle)