1 Calcular ∫ Sec(θ)dθ tenemos que ∫ Sec(θ)dθ = ∫ Sec(θ)( Sec(θ ...
1 Calcular ∫ Sec(θ)dθ tenemos que ∫ Sec(θ)dθ = ∫ Sec(θ)( Sec(θ ...
1 Calcular ∫ Sec(θ)dθ tenemos que ∫ Sec(θ)dθ = ∫ Sec(θ)( Sec(θ ...
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2<br />
<br />
<br />
= T an(<strong>θ</strong>)<strong>Sec</strong>(<strong>θ</strong>)T an(<strong>θ</strong>)d<strong>θ</strong> = T an 2 <br />
(<strong>θ</strong>)<strong>Sec</strong>(<strong>θ</strong>)d<strong>θ</strong> = (<strong>Sec</strong> 2 (<strong>θ</strong>)−1)<strong>Sec</strong>(<strong>θ</strong>)d<strong>θ</strong><br />
<br />
=<br />
<strong>Sec</strong> 3 (<strong>θ</strong>)−<strong>Sec</strong>(<strong>θ</strong>)d<strong>θ</strong> = 1<br />
<strong>Sec</strong>(<strong>θ</strong>)T an(<strong>θ</strong>)+1 Log(<strong>Sec</strong>(<strong>θ</strong>)+T an(<strong>θ</strong>))−Log(<strong>Sec</strong>(<strong>θ</strong>)+T an(<strong>θ</strong>))<br />
2 2<br />
= 1<br />
1<br />
1<br />
<strong>Sec</strong>(<strong>θ</strong>)T an(<strong>θ</strong>)− Log(<strong>Sec</strong>(<strong>θ</strong>)+T an(<strong>θ</strong>)) =<br />
2 2 2 (x√x2 − 1)− 1<br />
2 (x+√x2 − 1)<br />
En la hipérbola equilatera x 2 − y 2 = 1 definimos c = Cosh(x) s = Senh(x)<br />
y como c,s estan sobre la hipérbola c 2 − s 2 = 1<br />
c √ √ √<br />
Area = ”x” = sc − 2 x2 − 1dx = sc − c c2 − 1 + Log(c + c2 − 1)<br />
Por lo tanto<br />
1<br />
= sc − cs + Log(c + √ c 2 − 1) = Log(c + √ c 2 − 1)<br />
x = Log(c+ √ c 2 − 1) ⇒ e x = c+ √ c 2 − 1 ⇒ e x −c = √ c 2 − 1 ⇒ e 2x −2e x c+c 2 = c 2 −1<br />
⇒ e 2x − 2e x c = −1 ⇒ e 2x + 1 = 2e x c ⇒ e2x + 1<br />
2e x<br />
= c ⇒ c = ex + e −x<br />
De la relación c2 − s2 = 1 podemos obtener el valor de s<br />
s = √ c2 <br />
− 1 = ( ex + e−x )<br />
2<br />
2 <br />
e2x + 2 + e−2x e2x + 2 + e−2x − 4<br />
− 1 =<br />
− 1 =<br />
4<br />
4<br />
<br />
(ex − e−x ) 2<br />
=<br />
=<br />
4<br />
ex − e−x 2<br />
Tenemos <strong>que</strong><br />
2