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Which of the following integrals cannot be evaluated using a simple substitution? (4 points) Select one:a. the integral of the square root of the quantity x minus 1, dxb. the integral of the quotient of 1 and the square root of the quantity 1 minus x squared, dxc. the integral of the quotient of 1 and the square root of the quantity 1 minus x squared, dxd. the integral of x times the square root of the quantity x squared minus 1, dx

Which of the following integrals cannot be evaluated using a simple substitution 4 points Select onea the integral of the square root of the quantity x minus 1 class=

Respuesta :

Given:

There are given that the integral:

[tex]\int_{-8}^8\sqrt{64-x^2}dx[/tex]

Explanation:

From the given integral;

[tex]\int_{-8}^8\sqrt{64-x^2}dx[/tex]

Now,

Apply the trigonometry rule that is:

[tex]\int_{-a}^a\sqrt{b-x^2}dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}bcos^2(u)du[/tex]

Then,

From the function:

[tex]\int_{-8}^8\sqrt{64-x^2}dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}64cos^2udu[/tex]

Then,

Take the constant out:

So,

[tex]\begin{gathered} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}64cos^2udu=64\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}cos^2udu \\ =64\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1+cos2u}{2}du \\ =\frac{64}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}1+cos2udu \end{gathered}[/tex]

Then,

[tex]\begin{gathered} \frac{64}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}1+cos2udu=32\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}1du+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}cos2udu \\ =32(\pi+0) \\ =32\pi \end{gathered}[/tex]

Final answer:;

Hence, the correct option is C.

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