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

  • Thread starter mandymandy
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  • #1
Use integration to find the area of the region bounded by the given polar curves

r = [itex]\frac{3}{(1+cos \theta )}[/itex]

and

[itex]\theta[/itex] = [itex]\frac{\pi}{2}[/itex]



A = [itex]\frac{1}{2}[/itex] [itex]\int[/itex]f([itex]\theta[/itex])[itex]^{2}[/itex]d[itex]\theta[/itex]



My attempt:

(from -[itex]\frac{-\pi}{2}[/itex] to [itex]\frac{\pi}{2}[/itex] )

A = [itex]\frac{1}{2}[/itex][itex]\int[/itex] ([itex]\frac{3}{(1+cos \theta )}[/itex])[itex]^{2}[/itex] d[itex]\theta[/itex]

A = [itex]\frac{9}{2}[/itex][itex]\int[/itex] ([itex]\frac{1}{(1+cos \theta )}[/itex])[itex]^{2}[/itex] d[itex]\theta[/itex]

A = [itex]\frac{9}{2}[/itex][itex]\int[/itex] [itex]\frac{1}{(1+cos \theta )^{2}}[/itex])[itex][/itex] d[itex]\theta[/itex]

→[itex](1+cos \theta )^{2} [/itex] = [itex] cos^{2}\theta + 2cos\theta + 1 [/itex]

= 1 - sin[itex]^{2}[/itex][itex]\theta[/itex] + 2cos[itex]\theta[/itex] + 1

= 2 + 2cos [itex]\theta[/itex] - 1/2 + 1/2 cos 2 [itex]\theta[/itex]
...?
 

Answers and Replies

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

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