- #1
kirakun
- 25
- 2
Homework Statement
Well this is part of an integration process, namely:
[tex] \int \frac {sin^2x}{4+3cos^2x}dx [/tex]
Homework Equations
My attempt involved using a u-substitution, namely t = tan x
The Attempt at a Solution
Using t = tan x, [tex] sin^2 x = \frac {t^2}{1+t^2} [/tex] and [tex]cos^2 x = \frac {1}{1+t^2}[/tex]
we have [tex] \int \frac {sin^2x}{4+3cos^2x}dx [/tex]
which becomes
1. [tex]\int \frac {t^2}{(1+t^2)(4+ \frac {3}{1+t^2})} \times \frac {1}{1+t^2} [/tex]
2. [tex] \int \frac {t^2}{(1+t^2)( \frac {4+4t^2+3}{1+t^2})} \times \frac {1}{1+t^2} [/tex]
and finally yields
[tex] \int \frac {t^2}{(7+4t^2)(1+t^2)}dt [/tex]
So i am basically stuck at this point and cannot proceed further.
Maybe i should have used another substitution or I started it wrongly.
I cannot seem to get to the solution though i suspect arctan might be present in the answer.
Please do guide me.
Thanks again.
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