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

How to integrate dx/(4-5 sin x) using t-substitution method(i.e. taking tan x/2=t)?

- Calculus -
**Count Iblis**, Thursday, October 31, 2013 at 11:27am
sin(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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