Hi
Can someone tell me how i would integrate the following question:
1) $\displaystyle \int tan^2(x)$
P.S
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
$\displaystyle \chi$ $\displaystyle \sigma$
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
$\displaystyle \chi$ $\displaystyle \sigma$
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.