1. ## Simple Integration problem

Hi

Can someone tell me how i would integrate the following question:

1) $\displaystyle \int tan^2(x)$

P.S

2. Originally Posted by Paymemoney
Hi

Can someone tell me how i would integrate the following question:

1) $\displaystyle \int tan^2(x)$

P.S
$\displaystyle \int tan^2(x)dx$ = $\displaystyle \int (sec^2(x) - 1)dx$

3. Originally Posted by Paymemoney
1) $\displaystyle \int tan^2(x)$
$\displaystyle \int{\tan^2(x)}\;{dx} = \int{\left\{\sec^2(x)-1\right\}}\;{dx}$

EDIT: Bit late.

4. Because is...

$\displaystyle \tan^{2} x = \sin x \cdot \frac{\sin x}{\cos^{2} x}$ (1)

... You can integrate by parts and obtain...

$\displaystyle \int \tan^{2} x \cdot dx = \frac{\sin x}{\cos x} - \int dx = \tan x - x + c$ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. Can you show me how you have done it, because i still cannot get the right answer.

6. Originally Posted by Paymemoney
Can you show me how you have done it, because i still cannot get the right answer.

Do you know $\displaystyle \int\sec^2{x}dx$ ?

and $\displaystyle \int{dx}$ ?

7. Integration by parts:

$\displaystyle \int f(x)\cdot g(x)\cdot dx = f(x)\cdot G(x) - \int G(x)\cdot f^{'} (x)\cdot dx$ (1)

... where $\displaystyle G(*)$ is a primitive of $\displaystyle g(*)$. In our case is $\displaystyle f(x)= \sin x$ and $\displaystyle g(x)= \frac{\sin x}{\cos^{2} x}$ ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

8. Ok this is what i have done:

$\displaystyle \int tan^2(x) = \int sinx * \frac{sin}{cos^2(x)}$

$\displaystyle u= \frac{sin(x)}{cos^2(x)} du = \frac{cos^3(x) - 2cos(x)sin2(x)}{cos^4(x)}dx$

$\displaystyle dv = sin(x) v = -cos(x)$

$\displaystyle \frac{sin(x)}{cos^2(x)} * -cos(x) - \int \frac{cos^3(x) - 2cos(x)sin2(x)}{cos^4(x)} * -cos(x)$

$\displaystyle tan(x) + 2 \int cos^2(x)sin^2(x)$

Someone tell me where my mistake is.

9. Originally Posted by Paymemoney
Ok this is what i have done:

$\displaystyle \int tan^2(x) = \int sinx * \frac{sin}{cos^2(x)}$

$\displaystyle u= \frac{sin(x)}{cos^2(x)} du = \frac{cos^3(x) - 2cos(x)sin2(x)}{cos^4(x)}dx$

$\displaystyle dv = sin(x) v = -cos(x)$

$\displaystyle \frac{sin(x)}{cos^2(x)} * -cos(x) - \int \frac{cos^3(x) - 2cos(x)sin2(x)}{cos^4(x)} * -cos(x)$

$\displaystyle tan(x) + 2 \int cos^2(x)sin^2(x)$

Someone tell me where my mistake is.
you picked the wrong u and v

$\displaystyle u = \sin{x}$

$\displaystyle dv = \frac{\sin{x}}{\cos^2{x}}dx$

10. Originally Posted by 11rdc11
you picked the wrong u and v

$\displaystyle u = \sin{x}$

$\displaystyle dv = \frac{\sin{x}}{\cos^2{x}}dx$
how can i find what v is equal to?

11. By doing Integration by Substitution:

$\displaystyle \int \frac{\sin{x}}{\cos^2{x}}dx$

$\displaystyle u = \cos x \Longrightarrow du = -\sin x dx$

$\displaystyle = -\int \frac{1}{u^2}du$