Notes: TO calculate the continious integral of : 1/(1 + cos(theta))

Monday, October 20, 2014

TO calculate the continious integral of : 1/(1 + cos(theta))

You need to use the substitution $\displaystyle{\cos{\theta}}=\frac{{{1}-{{y}}^{{2}}}}{{{1}}}$
+ y^2)

$\displaystyle{y}={\tan{{\left(\frac{\theta}{{2}}\right)}}}=\gt{\left.{d}{y}\right.}={\left({d}\right.}$
theta)/(2cos^2(theta/2)) =gt d theta = (2(1 - y^2)dy)/(1 +
y^2)

$\displaystyle\int\frac{{{d}\theta}}{{{1}+{\cos{\theta}}}}=\int{\left({2}{\left({1}-\right.}\right.}$
y^2)dy)/(1 + y^2)/((1 + y^2 +1 - y^2)/(1 + y^2))

$\displaystyle\int$
(2(1 - y^2)dy)/(1 + y^2 +1 - y^2) = 2int ((1 -
y^2)dy)/2

$\displaystyle{2}\int\frac{{{\left({1}-{{y}}^{{2}}\right)}{\left.{d}{y}\right.}}}{{2}}=\int{\left.{d}{y}\right.}-\int{{y}}^{{2}}{\left.{d}{y}\right.}={y}$
- y^3/3 + c

You need to come back to the original variable
such that:

$\displaystyle\int\frac{{{d}\theta}}{{{1}+{\cos{\theta}}}}={\tan{{\left(\frac{\theta}{{2}}\right)}}}-$
(tan^3 (theta/2))/3 + c

Hence, evaluating
the given integral yields: $\displaystyle\int\frac{{{d}\theta}}{{{1}+{\cos{\theta}}}}={\tan{{\left(\frac{\theta}{{2}}\right)}}}-{\left({{\tan}}^{{3}}\right.}$
(theta/2))/3 + c

Notes

Monday, October 20, 2014

TO calculate the continious integral of : 1/(1 + cos(theta))

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What is the meaning of the 4th stanza of Eliot's Preludes, especially the lines "I am moved by fancies...Infinitely suffering thing".

Monday, October 20, 2014

TO calculate the continious integral of : 1/(1 + cos(theta))

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Post a Comment

What is the meaning of the 4th stanza of Eliot&#39;s Preludes, especially the lines &quot;I am moved by fancies...Infinitely suffering thing&quot;.

What is the meaning of the 4th stanza of Eliot's Preludes, especially the lines "I am moved by fancies...Infinitely suffering thing".