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February 27, 2015

February 27, 2015

Posted by **MS** on Thursday, October 31, 2013 at 8:15am.

- Calculus -
**Count Iblis**, Thursday, October 31, 2013 at 11:27amsin(x) = sin[2(x/2)] =

2 sin(x/2) cos(x/2)

Draw a right triangle with one angle equal to x/2. If you make the length of the side opposite to that angle equal to

t = tan(x/2) then the length of the side side orthogonal to it that connects to that angle will be equal to 1, because ratio of the two sides must be equal to tan(x/2) = t.

This means that the length of the hypotenuse is equal to:

h = sqrt(1+t^2)

and you have that:

sin(x/2) = t/sqrt(1+t^2)

cos(x/2) = 1/sqrt(1+t^2)

Therefore:

sin(x) = 2 t/(1+t^2)

And:

x = 2 arctan(t) ------>

dx = 2dt/(1+t^2)

This gives:

dx/(4-5 sin x)

2 dt/(1+t^2) 1/[4 - 10 t/(1+t^2)] =

dt/[2+2 t^2 -5t] =

dt/[2(t-1/2)(t-2)] =

dt/(t-1/2) * 1/[2* (1/2 - 2)] +

dt/[(t-2)] * 1/[2* (2-1/2)]

Integrating gives:

-1/3 Log|t-1/2| + 1/3 Log|t-2| =

1/3 Log|[tan(x/2) - 2]/[tan(x/2) -1/2]|

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