Question

Mais uma aí galera, quem puder me ajudar!

∫ x² -√1-x² dx

Answer 1

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


Substituição trigonométrica:

$x=\mathrm{sen\,}t~\Rightarrow~\left\{ \begin{array}{l} dx=\cos t\,dt\\ \\ t=\mathrm{arcsen}\,x~\Rightarrow~-\frac{\pi}{2}\leq t\leq \frac{\pi}{2} \end{array} \right.$


Então, temos que

$\sqrt{1-x^{2}}\\ \\ =\sqrt{1-\mathrm{sen^{2}\,}t}\\ \\ =\sqrt{\cos^{2}t}\\ \\ =\left|\cos t\right|$


Como $-\frac{\pi}{2}\leq t\leq \frac{\pi}{2}\,,$ segue que

$\cos t\geq 0~\Rightarrow~\left|\cos t|=\cos t.$


Substituindo, a integral fica

$I=\displaystyle\int{(\mathrm{sen^{2}\,}t-\cos t)\cos t\,dt}\\ \\ \\ =\int{\mathrm{sen^{2}\,}t\cos t\,dt}-\int{\cos^{2}t\,dt}\\ \\ \\ =\dfrac{(\mathrm{sen\,}t)^{3}}{3}-\underbrace{\int{\cos^{2}t\,dt}}_{I_{1}}~~~~~~\mathbf{(i)}$

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Vamos calcular $I_{1}$ utilizando a identidade trigonométrica

$\cos^{2}t=\dfrac{1}{2}+\dfrac{1}{2}\cos 2t$


Portanto,

$I_{1}=\displaystyle\int{\cos^{2}t\,dt}\\ \\ \\ =\int{\left(\dfrac{1}{2}+\dfrac{1}{2}\cos 2t \right )dt}\\ \\ \\ =\dfrac{1}{2}\int{dt}+\dfrac{1}{2}\int{\cos 2t\,dt}\\ \\ \\ =\dfrac{1}{2}\,t+\dfrac{1}{2}\cdot \left(\dfrac{1}{2}\,\mathrm{sen\,}2t \right )+C_{1}\\ \\ \\ =\dfrac{1}{2}\,t+\dfrac{1}{4}\,\mathrm{sen\,}2t+C_{1}\\ \\ \\ =\dfrac{1}{2}\,t+\dfrac{1}{4}\cdot (2\,\mathrm{sen\,}t\cos t)+C_{1}\\ \\ \\ =\dfrac{1}{2}\,t+\dfrac{1}{2}\,\mathrm{sen\,}t\cos t+C_{1}~~~~~~\mathbf{(ii)}$
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Substituindo $\mathbf{(ii)}$ em $\mathbf{(i)}\,,$ a integral fica

$I=\dfrac{(\mathrm{sen\,}t)^{3}}{3}-\left(\dfrac{1}{2}\,t+\dfrac{1}{2}\,\mathrm{sen\,}t\cos t \right )+C\\ \\ \\ =\dfrac{\mathrm{sen^{3}\,}t}{3}-\dfrac{1}{2}\,t-\dfrac{1}{2}\,\mathrm{sen\,}t\cos t+C\\ \\ \\ =\dfrac{x^{3}}{3}-\dfrac{1}{2}\,\mathrm{arcsen\,}x-\dfrac{1}{2}\,x\sqrt{1-x^{2}}+C$

$\therefore~\boxed{\begin{array}{c}\displaystyle\int{(x^{2}-\sqrt{1-x^{2}})\,dx}=\dfrac{x^{3}}{3}-\dfrac{1}{2}\,\mathrm{arcsen\,}x-\dfrac{1}{2}\,x\sqrt{1-x^{2}}+C \end{array}}$

Answer 2

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

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

$\frac{x^3}{3} - \int\ {\sqrt{1-x^2}}} \, dx$

$x=cos\theta$

$dx=-sin\theta d\theta$

$\frac{x^3}{3} - \int\ {-sin\theta\sqrt{1-cos^2\theta}}} \, d\theta$

$\frac{x^3}{3} + \int\ {sin\theta\sqrt{sin^2\theta}}} \, d\theta$

$\frac{x^3}{3} + \int\ {sin^2\theta} \, d\theta$

$sin^2\theta=\frac{1-cos(2\theta)}{2}$

$\frac{x^3}{3} + \int\ {\frac{1-cos(2\theta)}{2}} \, d\theta$

$\frac{x^3}{3} + \frac12\int\ {1-cos(2\theta)} \, d\theta$

$\frac{x^3}{3} + \frac12\int\ {1} \, d\theta - \frac12\int\ {cos(2\theta)} \, d\theta$

$\frac{x^3}{3} + \frac12\theta - \frac12\int\ {cos(2\theta)} \, d\theta$

$u=2\theta$

$d\theta=\frac{du}{2}$

$\frac{x^3}{3} + \frac12\theta - \frac12\int\ {\frac{cos(u)}{2}} \, du$

$\frac{x^3}{3} + \frac12\theta - \frac14\int\ {cos(u)} \, du$

$\frac{x^3}{3} + \frac12\theta - \frac14sinu+C$

$x=cos\theta$

$cos^{-1}x=\theta$

$\frac{x^3}{3} + \frac12cos^{-1}x - \frac14sinu+C$

$u=2\theta$

$\frac{x^3}{3} + \frac12cos^{-1}x - \frac14sin(2\theta)+C$

$sin(2\theta)=2sin\theta cos\theta$

$\frac{x^3}{3} + \frac12cos^{-1}x - \frac12sin\theta cos\theta+C$

$x=cos\theta$

$\frac{x^3}{3} + \frac12cos^{-1}x - \frac12sin\theta (x)+C$

Por pitagoras sabemos que:

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

$\boxed{\frac{x^3}{3} + \frac12cos^{-1}x - \frac12x\sqrt{1-x^2}+C}$