1. ## Rational Trignometric Integral

hmm any help?

2. Originally Posted by usagi_killer

hmm any help?
one of, i am sure, many ways:

multiply by $\displaystyle \frac {1 + \sin x}{1 + \sin x}$

you get

$\displaystyle \int \frac {(1 + \sin x)^2}{\cos^2 x}~dx$

Now expand the numerator and break this into 3 integrals. you should be able to handle each pretty easily

3. Originally Posted by usagi_killer

hmm any help?
Multiply the integrated function by $\displaystyle \frac{1+sin(x)}{1+sin(x)}$

With some algebra and trigonometric identities, you will get:
$\displaystyle \int \frac{sin^2(x)+2sin(x)+1}{cos^2(x)} dx$
$\displaystyle =\int tan^2(x)dx + 2 \int \frac{sin(x)}{cos^2(x)} dx + \int sec^2(x) dx$

For the first: use a well-known trigometric identity.
For the second: use a U-Substitution.
For the third: Its a well-known integral.

4. Originally Posted by usagi_killer

hmm any help?
$\displaystyle \int\frac{1+\sin(x)}{1-\sin(x)}\text{ }dx=\int\frac{1+\sin(x)}{1-\sin(x)}\cdot\frac{1+\sin(x)}{1+\sin(x)}\text{ }dx=$$\displaystyle \int\frac{\left(1+\sin(x)\right)^2}{1-\sin^2(x)}\text{ }dx=\int\frac{\left(1+\sin(x)\right)^2}{\cos^2(x)} \text{ }dx$. I'm sure you can go from there.

5. Originally Posted by General
Multiply the integrated function by $\displaystyle \frac{1+sin(x)}{1+sin(x)}$

With some algebra and trigonometric identities, you will get:
$\displaystyle \int \frac{sin^2(x)+2sin(x)+1}{cos^2(x)} dx$
$\displaystyle =\int tan^2(x)dx + 2 \int \frac{sin(x)}{cos^2(x)} dx + \int sec^2(x) dx$

For the first: use a well-known trigometric identity.
For the second: use a U-Substitution.
For the third: Its a well-known integral.
u-sub is not necessary for the second

6. Originally Posted by Jhevon
u-sub is not necessary for the second
But it solves it.