Question

(50 PONTOS) Sabendo que
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$\displaystyle\int\!\frac{1}{\sqrt{1-x^2}}\,dx=\mathrm{arcsen\,}x+C$

avalie a integral indefinida

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

sem recorrer à substituição trigonométrica.
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Answer 1

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

Fazendo

$u=\sqrt{1-x^{2}}~~\Rightarrow~~du=\dfrac{1}{2\sqrt{1-x^{2}}}(-2x)\,dx=\dfrac{-x}{\sqrt{1-x^{2}}}\,dx\\\\\\dv=dx~~\Rightarrow~~v=x$

e integrando por partes, temos

$\displaystyle\int\sqrt{1-x^{2}}\,dx=\int u\,dv=uv-\int v\,du\\\\\\=x\sqrt{1-x^{2}}-\int x\bigg[\dfrac{(-x)}{\sqrt{1-x^{2}}}\bigg]\,dx\\\\\\=x\sqrt{1-x^{2}}-\int\dfrac{-x^{2}}{\sqrt{1-x^{2}}}\,dx\\\\\\=x\sqrt{1-x^{2}}-\int\dfrac{1-x^{2}-1}{\sqrt{1-x^{2}}}\,dx\\\\\\=x\sqrt{1-x^{2}}+\int\dfrac{1}{\sqrt{1-x^{2}}}\,dx-\int\dfrac{1-x^{2}}{\sqrt{1-x^{2}}}\,dx\\\\\\=x\sqrt{1-x^{2}}+\arcsin(x)-\int\dfrac{1-x^{2}}{\sqrt{1-x^{2}}}\,dx$

Vamos avaliar essa integral:

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

Como a função está definida para $1-x^{2}$ estritamente positivo, podemos cancelar esse fator, ficando com

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

Logo, temos

$\displaystyle\int\sqrt{1-x^{2}}\,dx=x\sqrt{1-x^{2}}+\arcsin(x)-\int\dfrac{1-x^{2}}{\sqrt{1-x^{2}}}\,dx\\\\\\\int\sqrt{1-x^{2}}\,dx=x\sqrt{1-x^{2}}+\arcsin(x)-\int\sqrt{1-x^{2}}\,dx\\\\\\2\int\sqrt{1-x^{2}}\,dx=x\sqrt{1-x^{2}}+\arcsin(x)\\\\\\\boxed{\boxed{\int\sqrt{1-x^{2}}\,dx=\dfrac{x\sqrt{1-x^{2}}+\arcsin(x)}{2}+C}}$

Answer 2

$\displaystyle\sf{\int\sqrt{a^2-u^2}~du=\dfrac{u}{2}\sqrt{a^2-u^2}+\dfrac{a^2}{2}arcsen\left(\dfrac{u}{a}\right)+k}$

$\displaystyle\sf{\int\sqrt{1-x^2}dx=\dfrac{x}{2}\sqrt{1-x^2}+\dfrac{1}{2}arcsen(x)+c }$