# integration

• Sep 20th 2012, 10:20 PM
priya190992
integration
how do we integrate

(sin2 x)/cos x
• Sep 20th 2012, 10:42 PM
MaxJasper
Re: integration
Quote:

Originally Posted by priya190992
how do we integrate

(sin2 x)/cos x

Try: $\displaystyle \int \frac{\text{Sin}(x)^2}{\text{Cos}(x)} \, dx=\int -\frac{1}{2} (-1+\text{Cos}(2 x)) \text{Sec}(x)dx$
• Sep 21st 2012, 04:49 PM
johnsomeone
Re: integration
$\displaystyle \int \frac{\sin^2(x)}{\cos(x)} dx$

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

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

$\displaystyle = \int \frac{u^2}{1 - u^2} du$ (where $\displaystyle u = \sin(x)$).

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

$\displaystyle = \int \left( \frac{u^2 - 1}{1 - u^2} + \frac{1}{1 - u^2} \right) du$

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

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

$\displaystyle = -u + \int \frac{1}{(1 - u)(1+u)} du$

$\displaystyle = -u + \int \left( \frac{1/2}{(1 - u)} + \frac{1/2}{(1+u)} \right) du$

$\displaystyle = -u + (1/2) \int \frac{du}{(1 - u)} + (1/2) \int \frac{du}{(1+u)}$

$\displaystyle = -u + (1/2) (-\ln|1-u|) + (1/2) (\ln|1+u|) + C$

$\displaystyle = -u + (1/2) \ln\left( \left\lvert \frac{1+u}{1-u} \right\rvert \right) + C$

$\displaystyle = -\sin(x) + (1/2) \ln\left( \left\lvert \frac{1+\sin(x)}{1-\sin(x)} \right\rvert \right) + C$

$\displaystyle = -\sin(x) + (1/2) \ln\left( \left\lvert \frac{(1+\sin(x))^2}{\cos^2(x)} \right\rvert \right) + C$

$\displaystyle = -\sin(x) + (1/2) \ln\left( \left\lvert \frac{1+\sin(x)}{\cos(x)} \right\rvert^2 \right) + C$

$\displaystyle = -\sin(x) + \ln\left( \left\lvert \frac{1+\sin(x)}{\cos(x)} \right\rvert \right) + C$.

$\displaystyle = -\sin(x) + \ln\left( \frac{1+\sin(x)}{|\cos(x)|} \right) + C$.

*** Check ***

On a domain where $\displaystyle \cos(x) > 0$:

$\displaystyle \frac{d}{dx}\left\{ -\sin(x) + \ln\left( \frac{(1+\sin(x))}{\cos(x)} \right) + C \right\}$

$\displaystyle = -(\cos(x)) + \frac{\cos(x)}{(1+\sin(x))} \frac{d}{dx} \left\{ \frac{(1+\sin(x))}{\cos(x)} \right\} + (0)$

$\displaystyle = -\cos(x) + \frac{\cos(x)}{(1+\sin(x))} \left( \frac{(\cos(x))(\cos(x))-(1+\sin(x))(-\sin(x))}{\cos^2(x)} \right)$

$\displaystyle = -\cos(x) + \frac{\cos(x)}{(1+\sin(x))} \left( \frac{\cos^2(x)+\sin^2(x) + \sin(x)}{\cos^2(x)} \right)$

$\displaystyle = -\cos(x) + \frac{\cos(x)}{(1+\sin(x))} \left( \frac{1 + \sin(x)}{\cos^2(x)} \right)$

$\displaystyle = -\cos(x) + \frac{1}{\cos(x)} = -\frac{\cos^2(x)}{\cos(x)} + \frac{1}{\cos(x)} = \frac{1 - \cos^2(x)}{\cos(x)}$

$\displaystyle = \frac{\sin^2(x)}{\cos(x)}$.
• Sep 25th 2012, 06:50 PM
Soroban
Re: integration
Hello, priya190992!

Quote:

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

We have: .$\displaystyle \displaystyle\int\frac{1-\cos^2\!x}{\cos x}\,dx \;=\;\int\left(\frac{1}{\cos x} - \cos x\right)dx \;=\;\int(\sec x - \cos x)\,dx$

. . . $\displaystyle =\;\ln|\sec x + \tan x| - \sin x + C$