Question

Calcule a integral indefinida

$\mathsf{\displaystyle\int~\sqrt{-x^2+4x+5}~dx}$

Por favor responder de forma detalhada.

Answer 1

É dada a integral indefinida:

$\displaystyle I=\int\sqrt{-x^2+4x+5}\,dx$

Vamos desenvolver a expressão que está dentro da raiz quadrada:

$-x^2+4x+5=-x^2+4x-4+4+5\\\\ -x^2+4x+5=-(x-2)^2+9\\\\ -x^2+4x+5=9-(x-2)^2\\\\ -x^2+4x+5=9\left(1-\left(\dfrac{x-2}{3}\right)^2\right)$

Desse modo, a integral que queremos calcular equivale a:

$\displaystyle I=\int\sqrt{-x^2+4x+5}\,dx\\\\ I=\int\sqrt{9\left(1-\left(\dfrac{x-2}{3}\right)^2\right)}\,dx\\\\ I=\int3\sqrt{1-\left(\dfrac{x-2}{3}\right)^2}\,dx$

O termo dentro da raiz quadrada deve ser não-negativo. Logo:

$1-\left(\dfrac{x-2}{3}\right)^2\ge0\\\\ \left(\dfrac{x-2}{3}\right)^2\le1\\\\ -1\le \dfrac{x-2}{3}\le1$

Pelo intervalo obtido, podemos fazer a seguinte substituição:

$\dfrac{x-2}{3}=\sin(\theta)\\\\ x=3\sin(\theta)+2\\\\ dx=3\cos(\theta)\,d\theta$

Utilizando na integral dada:

$\displaystyle I=\int3\sqrt{1-\left(\dfrac{x-2}{3}\right)^2}\,dx\\\\ I=3\int\sqrt{1-\sin^2(\theta)}\,3\cos(\theta)\,d\theta\\\\ I=9\int \cos^2(\theta)\,d\theta$

Pelas relações trigonométricas conhecidas:
$\cos(2\theta)=2\cos^2(\theta)-1\Longrightarrow \cos^2(\theta)=\dfrac{1}{2}\cos(2\theta)+\dfrac{1}{2}$

Aplicando em I:

$\displaystyle I=9\int \cos^2(\theta)\,d\theta\\\\ I=9\int\left(\dfrac{1}{2}\cos(2\theta)+\dfrac{1}{2}\right)\,d\theta\\\\ I=\dfrac{9}{2}\int \cos(2\theta)\,d\theta+\dfrac{9}{2}\int d\theta\\\\ I=\dfrac{9}{2}\cdot\dfrac{1}{2}\sin(2\theta)+\dfrac{9}{2}\theta+C\\\\ I=\dfrac{9}{4}\sin(2\theta)+\dfrac{9}{2}\theta+C$

Voltando à variável $x$:

$I=\dfrac{9}{4}\sin(2\theta)+\dfrac{9}{2}\theta+C\\\\ I=\dfrac{9}{4}\cdot2\cdot\sin(\theta)\cdot\cos(\theta)+\dfrac{9}{2}\theta+C\\\\ I=\dfrac{9}{2}\sin(\theta)\cdot\cos(\theta)+\dfrac{9}{2}\theta+C\\\\ I=\dfrac{9}{2}\cdot\left(\dfrac{x-2}{3}\right)\cdot\cos\left(\arcsin\left(\dfrac{x-2}{3}\right)\right)+\dfrac{9}{2}\arcsin\left(\dfrac{x-2}{3}\right)+C\\\\ I=\dfrac{3}{2}(x-2)\cdot\sqrt{1-\left(\dfrac{x-2}{3}\right)^2}+\dfrac{9}{2}\arcsin\left(\dfrac{x-2}{3}\right)+C$
$I=\dfrac{3}{2}(x-2)\cdot\sqrt{1-\left(\dfrac{x^2-4x-4}{9}\right)}+\dfrac{9}{2}\arcsin\left(\dfrac{x-2}{3}\right)+C\\\\ I=\dfrac{3}{2}(x-2)\cdot\dfrac{1}{3}\sqrt{9-(x^2-4x-4)}+\dfrac{9}{2}\arcsin\left(\dfrac{x-2}{3}\right)+C\\\\ I=\dfrac{1}{2}(x-2)\sqrt{-x^2+4x+5}+\dfrac{9}{2}\arcsin\left(\dfrac{x-2}{3}\right)+C$
$\displaystyle \boxed{\int\sqrt{-x^2+4x+5}\,dx=\dfrac{x-2}{2}\sqrt{-x^2+4x+5}+\dfrac{9}{2}\arcsin\left(\dfrac{x-2}{3}\right)\!+\!C}$