# What is the integral of 1/((1+cosx)^2)?

1. Feb 2, 2013

### mandymandy

Use integration to find the area of the region bounded by the given polar curves

r = $\frac{3}{(1+cos \theta )}$

and

$\theta$ = $\frac{\pi}{2}$

A = $\frac{1}{2}$ $\int$f($\theta$)$^{2}$d$\theta$

My attempt:

(from -$\frac{-\pi}{2}$ to $\frac{\pi}{2}$ )

A = $\frac{1}{2}$$\int$ ($\frac{3}{(1+cos \theta )}$)$^{2}$ d$\theta$

A = $\frac{9}{2}$$\int$ ($\frac{1}{(1+cos \theta )}$)$^{2}$ d$\theta$

A = $\frac{9}{2}$$\int$ $\frac{1}{(1+cos \theta )^{2}}$) d$\theta$

→$(1+cos \theta )^{2}$ = $cos^{2}\theta + 2cos\theta + 1$

= 1 - sin$^{2}$$\theta$ + 2cos$\theta$ + 1

= 2 + 2cos $\theta$ - 1/2 + 1/2 cos 2 $\theta$
...?

2. Feb 2, 2013

### LCKurtz

Perhaps you can use $1+\cos\theta = 2\cos^2(\frac\theta 2)$.