Question

calculo II
∫x³-√1-x²dx

Answer 1

$\displaystyle\int x^{3}-\sqrt{1-x^{2}}dx=\int x^{3}dx-\int\sqrt{1-x^{2}}dx\\\\\\\displaystyle\int x^{3}-\sqrt{1-x^{2}}dx=\dfrac{x^{4}}{4}+c_{1}-\int\sqrt{1-x^{2}}dx$
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Fazendo $x=sen(\theta),~~\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]$, temos $dx=cos(\theta)d\theta$, então

$\displaystyle\int\sqrt{1-x^{2}}dx=\int\sqrt{1-(cos(\theta))^{2}}cos(\theta)d\theta\\\\\\\displaystyle\int\sqrt{1-x^{2}}dx=\int\sqrt{1-sen^{2}(\theta)}cos(\theta)d\theta$

Da relação fundamental da trigonometria, tiramos que $1-sen^{2}(\theta)=cos^{2}(\theta)$

$\displaystyle\int\sqrt{1-x^{2}}dx=\int\sqrt{cos^{2}(\theta)}cos(\theta)d\theta\\\\\\\int\sqrt{1-x^{2}}dx=\int|cos(\theta)|cos(\theta)d\theta$

Como $\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]$, cos(θ) ≥ 0, então |cos(θ)| = cos(θ)

$\displaystyle\int\sqrt{1-x^{2}}dx=\int cos(\theta)cos(\theta)d\theta=\int cos^{2}(\theta)d\theta$

Note que

$cos(2\theta)=cos(\theta+\theta)=cos^{2}(\theta)-sen^{2}(\theta)\\\\cos(2\theta)=cos^{2}(\theta)-[1-cos^{2}(\theta)]\\\\cos(2\theta)=cos^{2}(\theta)-1+cos^{2}(\theta)\\\\2cos^{2}(\theta)=cos(2\theta)+1\\\\\\\boxed{\boxed{cos^{2}(\theta)=\dfrac{cos(2\theta)+1}{2}}}$

Então:

$\displaystyle\int\sqrt{1-x^{2}}dx=\int\dfrac{cos(2\theta)+1}{2}d\theta\\\\\\\int\sqrt{1-x^{2}}dx=\dfrac{1}{2}\int cos(2\theta)d\theta+\dfrac{1}{2}\int d\theta\\\\\\\int\sqrt{1-x^{2}}dx=\dfrac{1}{2}\int cos(2\theta)d\theta+\dfrac{\theta}{2}$

Fazendo $v=2\theta~\longrightarrow~dv=2d\theta~\longrightarrow~d\theta=\frac{1}{2}dv$

$\displaystyle\int\sqrt{1-x^{2}}dx=\dfrac{\theta}{2}+\dfrac{1}{2}\int cos(v)\dfrac{1}{2}dv\\\\\\\int\sqrt{1-x^{2}}dx=\dfrac{\theta}{2}+\dfrac{1}{4}\int cos(v)dv\\\\\\\int\sqrt{1-x^{2}}dx=\dfrac{\theta}{2}+\dfrac{1}{4}sen(v)+C\\\\\\\int\sqrt{1-x^{2}}dx=\dfrac{\theta}{2}+\dfrac{sen(2\theta)}{4}+C$

Voltando para x:

$x=sen(\theta)~~~\therefore~~~\boxed{\boxed{\theta=arcsen(x)}}\\\\\\x=sen(\theta)~~\therefore~~2xcos(\theta)=2sen(\theta)cos(\theta)=sen(2\theta)$

Mas, como $cos(\theta)=\sqrt{1-sen^{2}(\theta)}=\sqrt{1-x^{2}}$, temos

$sen(2\theta)=2x\sqrt{1-x^{2}}$

Finalmente:

$\displaystyle\int\sqrt{1-x^{2}}dx=\dfrac{\theta}{2}+\dfrac{sen(2\theta)}{2}+c_{2}\\\\\\\int\sqrt{1-x^{2}}dx=\dfrac{arcsen(x)}{2}+\dfrac{2x\sqrt{1-x^{2}}}{4}+c_{2}\\\\\\\boxed{\boxed{\int\sqrt{1-x^{2}}dx=\dfrac{arcsen(x)}{2}+\dfrac{x\sqrt{1-x^{2}}}{2}+c_{2}}}$
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Então:

$\displaystyle\int x^{3}-\sqrt{1-x^{2}}dx=\dfrac{x^{4}}{4}+c_{1}-\int\sqrt{1-x^{2}}dx\\\\\\\int x^{3}-\sqrt{1-x^{2}}dx=\dfrac{x^{4}}{4}+c_{1}-\dfrac{arcsen(x)}{2}-\dfrac{x\sqrt{1-x^{2}}}{2}-c_{2}\\\\\\\boxed{\boxed{\int x^{3}-\sqrt{1-x^{2}}dx=\dfrac{x^{4}}{4}-\dfrac{arcsen(x)}{2}-\dfrac{x\sqrt{1-x^{2}}}{2}+C}}$