Integral rational trigonometric
1 answer:
Substitute <em>x</em> = 3 - 2 cos(<em>θ</em>) and d<em>x</em> = 2 sin(<em>θ</em>) d<em>θ</em> (where "sin" = "sen"). So we have
∫ sin(<em>θ</em>) / (3 - 2 cos(<em>θ</em>)) d<em>θ</em> = 1/2 ∫ 1/<em>x</em> d<em>x</em>
= 1/2 ln|<em>x</em>| + <em>C</em>
= 1/2 ln(3 - 2 cos( <em>θ</em> )) + <em>C</em>
(We can remove the absolute value because -1 ≤ cos(<em>θ</em>) ≤ 1, so 1 ≤ 3 - 2 cos(<em>θ</em>) ≤ 5, and |<em>x</em>| = <em>x</em> when <em>x</em> ≥ 0.)
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