9:27 PM Mon 17 Mar . 90\% mylab.pearson.com Kurulus Osm... Can't Open P... (c) CUNY Login The Evolutio... Discord Do Homewor... MAT 220-S25 Rasheed Larry 03/17/25 9:27 PM HW Score: 33.33\%, 1 Homework: Integration using Tables Question 2, 8.6.31 of 3 points ○ Points: 0 of 1 Save Question list Question 1 Question 2 Question 3 Choose the correct answer below. A. $$ \frac{1}{15}\left(\sin ^{-1} \sqrt{15 x}-\sqrt{15-225 x^{2}}\right)+C $$ B. $$ \frac{1}{15}\left(\sin ^{-1} \sqrt{1-15 x}-\sqrt{15 x-225 x^{2}}\right)+C $$ C. $$ \frac{1}{15}\left(\sin ^{-1} \sqrt{15 x}-\sqrt{15 x-225 x^{2}}\right)+C $$ D. $$ \frac{1}{15}\left(\cos ^{-1} \sqrt{15 x}-\sqrt{15 x-225 x^{2}}\right)+C $$ Help me solve this View an example Get more help • Clear all Check answer

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UpStudy AI · Accepted Answer

We start by noting that the answer choices have two common ingredients: an inverse‐trigonometric term and a square root term. All four answers are of the form

$$
\frac{1}{15}\Bigl(\, \text{inverse trig function of a square root} - \sqrt{\phantom{X}}\Bigr)+C.
$$

A very common integration result (found in many integration tables) is one for an integral of the form

$$
\int \frac{dx}{\sqrt{15x-225x^2}}
$$

or, after factoring the radicand,
 
$$
15x-225x^2=15x(1-15x).
$$

A standard antiderivative obtained via a table (or by using the substitution
$$
u=\sqrt{15x}
$$
so that after some algebra the answer involves an inverse sine whose argument is $$\sqrt{15x}$$ and a square root like $$\sqrt{15x-225x^2}=\sqrt{15x(1-15x)}$$) is shown by the option

$$
\frac{1}{15}\left(\sin ^{-1} \sqrt{15x}-\sqrt{15x-225x^2}\right)+C.
$$

Notice that among the four answer choices only option C displays both the inverse sine and the radicand in a consistent way. In option A the radical in the second term is $$\sqrt{15-225x^2}$$, which does not match the factorization; option B has a different argument (and the wrong sign inside the square root); and option D uses an inverse cosine instead of an inverse sine.

Thus, following the structure expected from using integration tables, the correct antiderivative is

$$
\frac{1}{15}\left(\sin ^{-1} \sqrt{15x}-\sqrt{15x-225x^2}\right)+C.
$$

Therefore, the correct answer is option C.
The correct answer is option C: $$ \frac{1}{15}\left(\sin ^{-1} \sqrt{15x}-\sqrt{15x-225x^{2}}\right)+C $$.

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