Question

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

Answer 1

Given:

There are given that the integral:

`\int_(-8)^8√(64-x^2)dx`

Step-by-step explanation:

From the given integral;

`\int_(-8)^8√(64-x^2)dx`

Now,

Apply the trigonometry rule that is:

`\int_(-a)^a√(b-x^2)dx=\int_{-(\pi)/(2)}^{(\pi)/(2)}bcos^2(u)du`

Then,

From the function:

`\int_(-8)^8√(64-x^2)dx=\int_{-(\pi)/(2)}^{(\pi)/(2)}64cos^2udu`

Then,

Take the constant out:

So,

`\begin{gathered} \int_{-(\pi)/(2)}^{(\pi)/(2)}64cos^2udu=64\int_{-(\pi)/(2)}^{(\pi)/(2)}cos^2udu \\ =64\int_{-(\pi)/(2)}^{(\pi)/(2)}(1+cos2u)/(2)du \\ =(64)/(2)\int_{-(\pi)/(2)}^{(\pi)/(2)}1+cos2udu \end{gathered}`

Then,

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

Final answer:;

Hence, the correct option is C.