Question

∫√(1 -x²)dx ; É uma integral definida, na qual o b = 1 e a =0.

Answer 1

$\displaystyle\int\limits_{0}^{1}{\sqrt{1-x^{2}}\,dx}$


Substituição trigonométrica:

$x=\mathrm{sen\,}\theta\;\;\;\Rightarrow\;\;\left\{ \begin{array}{l} dx=\cos \theta\,d \theta\\ \\ \theta=\mathrm{arcsen\,}x,\;\;\;-\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2} \end{array} \right.\\ \\ \\ \sqrt{1-x^{2}}=\sqrt{1-\mathrm{sen^{2}\,}\theta}\\ \\ \sqrt{1-x^{2}}=\sqrt{\cos^{2}\theta}\\ \\ \sqrt{1-x^{2}}=|\cos \theta|$


Como $\theta$ está no intervalo $\left[-\frac{\pi}{2},\;\frac{\pi}{2} \right ],$

$\cos \theta$ nunca é negativo, e temos então que


$|\cos \theta|=\cos \theta\\ \\ \sqrt{1-x^{2}}=\cos \theta$


Trocando os limites de integração:

$x=0\;\;\Rightarrow\;\;\theta=\mathrm{arcsen\,}0=0\\ \\ x=1\;\;\Rightarrow\;\;\theta=\mathrm{arcsen\,}1=\frac{\pi}{2}$


Substituindo na integral:

$\displaystyle\int\limits_{0}^{1}{\sqrt{1-x^{2}}\,dx}\\ \\ \\ =\displaystyle\int\limits_{0}^{\pi/2}{\cos \theta\cdot \cos \theta \,d \theta}\\ \\ \\ =\displaystyle\int\limits_{0}^{\pi/2}{\cos^{2}\theta\,d \theta}\\ \\ \\ =\displaystyle\int\limits_{0}^{\pi/2}{\dfrac{1+\cos 2\theta}{2}\,d \theta}\\ \\ \\ =\dfrac{1}{2}\displaystyle\int\limits_{0}^{\pi/2}{d \theta}+\dfrac{1}{2}\displaystyle\int\limits_{0}^{\pi/2}{\cos 2\theta\,d \theta}$


$=\dfrac{1}{2}\displaystyle\int\limits_{0}^{\pi/2}{d \theta}+\dfrac{1}{2}\displaystyle\int\limits_{0}^{\pi/2}{\cos 2\theta\,d \theta}\\ \\ \\ =\dfrac{1}{2}\left[\theta \right ]_{0}^{\pi/2}+\dfrac{1}{2}\cdot \left[\dfrac{\mathrm{sen\,}2\theta}{2} \right ]_{0}^{\pi/2}\\ \\ \\ =\dfrac{1}{2}\cdot \left[\dfrac{\pi}{2}-0 \right ]+\dfrac{1}{2}\cdot \left[\dfrac{\mathrm{sen\,}\pi}{2}-\dfrac{\mathrm{sen\,}0}{2} \right ]\\ \\ \\ =\dfrac{\pi}{4}+\dfrac{1}{2}\cdot 0\\ \\ \\ =\dfrac{\pi}{4}$