Calcule:
\int_{0}^{\sqrt{3} / 2}\left|\frac{1}{\sqrt{1-x^{2}}}-\sqrt{2}\right| d x

Passo 1

O enunciado pede
\int_{0}^{\sqrt{3}/2}\left|\frac{1}{\sqrt{1-x^{2}}}-\sqrt{2}\right|dx.
Vamos usar duas ideias principais: (i) achar o ponto onde o que está dentro do valor absoluto muda de sinal, resolvendo \tfrac{1}{\sqrt{1-x^2}}=\sqrt{2}, e (ii) saber que
\int \frac{dx}{\sqrt{1-x^2}}=\arcsin(x).

Passo 2

Resolvendo
\frac{1}{\sqrt{1-x^2}}=\sqrt{2} \quad\Longrightarrow\quad \sqrt{1-x^2}=\frac1{\sqrt2} \quad\Longrightarrow\quad 1-x^2=\frac12 \quad\Longrightarrow\quad x^2=\frac12 \quad\Longrightarrow\quad x=\frac1{\sqrt2}.
Denotamos a=\tfrac1{\sqrt2} e b=\tfrac{\sqrt3}{2}.

Passo 3

Para 0\le x<a temos \tfrac1{\sqrt{1-x^2}}<\sqrt2, logo \bigl|\tfrac1{\sqrt{1-x^2}}-\sqrt2\bigr|=\sqrt2-\tfrac1{\sqrt{1-x^2}}.
Para a<x\le b o sinal se inverte e \bigl|\tfrac1{\sqrt{1-x^2}}-\sqrt2\bigr|=\tfrac1{\sqrt{1-x^2}}-\sqrt2.
Assim o integral se divide em
\int_{0}^{b}\bigl|\dots\bigr|dx = \int_{0}^{a}\Bigl(\sqrt2-\frac1{\sqrt{1-x^2}}\Bigr)dx + \int_{a}^{b}\Bigl(\frac1{\sqrt{1-x^2}}-\sqrt2\Bigr)dx.

Passo 4

Calcular
I_1=\int_{0}^{a}\Bigl(\sqrt2-\frac1{\sqrt{1-x^2}}\Bigr)dx =\sqrt2\!\int_{0}^{a}\!dx -\int_{0}^{a}\frac{dx}{\sqrt{1-x^2}}.
A primeira parte é \sqrt2\cdot a=\sqrt2\cdot\frac1{\sqrt2}=1.
A segunda parte é \bigl[\arcsin(x)\bigr]_{0}^{a}=\arcsin\!\bigl(\tfrac1{\sqrt2}\bigr)-0=\frac\pi4.
Logo
I_1 = 1 - \frac\pi4.

Passo 5

Calcular
I_2=\int_{a}^{b}\Bigl(\frac1{\sqrt{1-x^2}}-\sqrt2\Bigr)dx =\int_{a}^{b}\frac{dx}{\sqrt{1-x^2}} -\sqrt2\!\int_{a}^{b}\!dx.
A primeira parte é \bigl[\arcsin(x)\bigr]_{a}^{b} =\arcsin\!\bigl(\tfrac{\sqrt3}{2}\bigr)-\arcsin\!\bigl(\tfrac1{\sqrt2}\bigr) =\frac\pi3-\frac\pi4=\frac\pi{12}.
A segunda parte é \sqrt2\,(b-a) =\sqrt2\Bigl(\frac{\sqrt3}{2}-\frac1{\sqrt2}\Bigr) =\frac{\sqrt6}{2}-1.
Portanto
I_2 = \frac\pi{12} \;-\;\Bigl(\frac{\sqrt6}{2}-1\Bigr) =\frac\pi{12}-\frac{\sqrt6}{2}+1.

Passo 6

Somando
I = I_1 + I_2 =\Bigl(1 - \tfrac\pi4\Bigr) +\Bigl(\tfrac\pi{12}-\tfrac{\sqrt6}{2}+1\Bigr) =2 - \frac\pi6 - \frac{\sqrt6}{2}.

Resposta

2 \;-\;\frac{\pi}{6}\;-\;\frac{\sqrt{6}}{2}.

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