Question

integral raiz quadrada 1-4x^2. Resolver por substituição trigonométrica

Answer 1

É dada a integral indefinida:

$\displaystyle I= \int \sqrt{1-4x^2}\,dx$

Como o integrando apresenta uma expressão dentro de uma raiz quadrada, podemos dizer que ela é não-negativa. Assim:

$1-4x^2\ge 0 \\\\ 4x^2\le 1\\\\ (2x)^2\le 1\\\\ -1\le2x\le1$

Como $-1\le2x\le1$, podemos fazer a substituição trigonométrica a seguir:

$2x=\sin(\theta),~~~\theta\in\left[\dfrac{-\pi}{2},\dfrac{\pi}{2}\right]\\\\ x=\dfrac{1}{2}\sin(\theta)\Longrightarrow \theta=\arcsin(2x)\\\\ dx=\dfrac{1}{2}\cos(\theta)\,d\theta$

Substituindo na integral dada:

$\displaystyle I= \int \sqrt{1-4x^2}\,dx\\\\ \displaystyle I= \int \sqrt{1-(2x)^2}\,dx\\\\ \displaystyle I= \int \sqrt{1-\sin^2(\theta)}\cdot \dfrac{1}{2}\cos(\theta)\,d\theta$

Usando que$\sin^2(\theta)+\cos^2(\theta)=1\Longrightarrow \sin^2(\theta)=1-\cos(\theta)$:

$\displaystyle I= \int \sqrt{1-\sin^2(\theta)}\cdot \dfrac{1}{2}\cos(\theta)\,d\theta\\\\ I= \int \sqrt{\cos^2(\theta)}\cdot \dfrac{1}{2}\cos(\theta)\,d\theta\\\\ I= \int \cos(\theta)\cdot \dfrac{1}{2}\cos(\theta)\,d\theta\\\\ I=\dfrac{1}{2}\int \cos^2(\theta)\,d\theta$

Agora, podemos utilizar que: $\cos(2\theta)=2\cos^2(\theta)-1\Longrightarrow \cos^2(\theta)=\dfrac{1}{2}\cos(2\theta)+\dfrac{1}{2}$:

$\displaystyle I=\dfrac{1}{2}\int \cos^2(\theta)\,d\theta\\\\ I=\dfrac{1}{2}\int \left(\dfrac{1}{2}\cos(2\theta)+\dfrac{1}{2}\right)\,d\theta\\\\ I=\dfrac{1}{4}\int\cos(2\theta)\,d\theta+\dfrac{1}{4}\int d\theta\\\\ I=\dfrac{1}{4}\cdot\left(\dfrac{1}{2}\sin(2\theta)\right)+\dfrac{1}{4}\theta+C\\\\ I=\dfrac{1}{8}\sin(2\theta)+\dfrac{1}{4}\theta+C$

Agora, podemos retornar à variável x:

$I=\dfrac{1}{8}\sin(2\theta)+\dfrac{1}{4}\theta+C\\\\ I=\dfrac{1}{8}\cdot2\sin(\theta)\cos(\theta)+\dfrac{1}{4}\theta+C\\\\ I=\dfrac{1}{8}\cdot2\sin(2\arcsin(2x))\cos(2\arcsin(2x))+\dfrac{1}{4}\arcsin(2x)+C\\\\ I=\dfrac{1}{8}\cdot2\cdot2x\cdot\sqrt{1-(2x)^2}+\dfrac{1}{4}\arcsin(2x)+C\\\\ I=\dfrac{1}{2}x\sqrt{1-4x^2}+\dfrac{1}{4}\arcsin(2x)+C\\\\ \boxed{\displaystyle \int \sqrt{1-4x^2}\,dx=\dfrac{1}{2}x\sqrt{1-4x^2}+\dfrac{1}{4}\arcsin(2x)+C}$

Answer 2

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$\Large\boxed{\begin{array}{c}\sf integral~envolvendo~\sqrt{a^2-u^2}\\\sf use~a~substituic_{\!\!,}\tilde ao~u=a~sen(\theta)\end{array}}$

$\displaystyle\sf\int\sqrt{1-4x^2}dx\\\sf fac_{\!\!,}a~2x=1sen(\theta)\implies x=\dfrac{1}{2}sen(\theta)\\\sf dx=\dfrac{1}{2}cos(\theta)d\theta\\\sf\sqrt{1-4x^2}=\sqrt{1-sen^2(\theta)}=\sqrt{cos^2(\theta)}=cos(\theta)\\\displaystyle\sf\int\sqrt{1-4x^2}dx=\int cos(\theta)\cdot\dfrac{1}{2}cos(\theta)d\theta=\dfrac{1}{2}\int cos^2(\theta)d\theta\\\boxed{\sf cos^2(\theta)=\dfrac{1}{2}[1+cos(2\theta)]}\\\displaystyle\sf\dfrac{1}{2}\int cos^2(\theta)d\theta=\dfrac{1}{2}\cdot\dfrac{1}{2}\int(1+cos(2\theta)d\theta$

$\displaystyle\sf \dfrac{1}{4}\int d\theta+\dfrac{1}{4}\int cos(2\theta)d\theta\\\sf \dfrac{1}{4}\theta+\dfrac{1}{4}\cdot\dfrac{1}{2}sen(2\theta)+k\\\sf \dfrac{1}{4}\theta+\dfrac{1}{4}\cdot\dfrac{1}{\diagup\!\!\!2}\cdot\diagup\!\!\!2\cdot sen(\theta)\cdot cos(\theta)=\dfrac{1}{4}\theta+\dfrac{1}{4}sen(\theta)cos(\theta)+k$

$\boxed{\begin{array}{c}\tt \theta=arc~sen(2x)\\\tt sen(\theta)=2x\\\tt cos(\theta)=\sqrt{1-4x^2}\end{array}}\\\displaystyle\sf\int\sqrt{1-4x^2}~dx=\dfrac{1}{4}arc~sen(2x)+\dfrac{1}{\diagdown\!\!\!\!\!4~_2}\cdot\diagdown\!\!\!\!2x\cdot\sqrt{1-4x^2}+k\\\boxed{\displaystyle\sf\int\sqrt{1-4x^2}~dx=\dfrac{1}{4}arc~sen(2x)+\dfrac{1}{2}x\sqrt{1-4x^2}+k}$