integration

**priya190992** · September 20th 2012, 10:20 PM

how do we integrate

(sin²x)/cos x

**MaxJasper** · September 20th 2012, 10:42 PM

Originally Posted by priya190992

how do we integrate

(sin²x)/cos x

Try: $\int \frac{\text{Sin}(x)^2}{\text{Cos}(x)} \, dx=\int -\frac{1}{2} (-1+\text{Cos}(2 x)) \text{Sec}(x)dx$

**johnsomeone** · September 21st 2012, 04:49 PM

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

*** Check ***

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

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

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

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

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

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

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

$= \frac{\sin^2(x)}{\cos(x)}$ .

**Soroban** · September 25th 2012, 06:50 PM

Hello, priya190992!

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

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

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

Thread: integration

LinkBack

Thread Tools

Display

integration

Re: integration

Re: integration

Re: integration

Similar Math Help Forum Discussions

Integration help please these are three small even basic integration problems

sketch the region of integration and change the order of integration.

Definite Integration - Double integration by parts

Simple Integration problem: Trigonometric Integration

Tricky integration, possibly Integration Factor

Search Tags