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

[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} \intf(\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
...?

 

[LCKurtz]

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