9:24 PM Mon 17 Mar . 91\% 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 9:24 PM HW Score: 0\%, 0 of 3 $$ \equiv $$ Homework: Integration Question 1, 8.6.29 points using Tables Points: 0 of 1 Save Question list Question 1 Question 2 Question 3 Use a substitution to change the following integral into one that can be found in the table of integrals. Then evaluate the integral. $$ \int \boldsymbol{\operatorname { s i n }}^{-1} \sqrt{x} d x $$ Which of the following is the value of the given integral? A. $$ \left(x-\frac{1}{2}\right) \sin ^{-1} \sqrt{x}+\frac{1}{2} \sqrt{x-x^{2}}+C $$ B. $$ \left(x+\frac{1}{2}\right) \sin ^{-1} \sqrt{x}+\frac{1}{2} \sqrt{x-x^{2}}+C $$ C. $$ \sqrt{x} \sin ^{-1} \sqrt{x}-\sqrt{1-x^{2}}+C $$ D. $$ \sqrt{x} \sin ^{-1} \sqrt{x}+\sqrt{1-x}+C $$ Textbook Calculator Clear all Check answer

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We start with the integral

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

and make the substitution

$$
u=\sqrt{x}\quad\Longrightarrow\quad x=u^2,\quad dx=2u\,du.
$$

Thus the integral becomes

$$
\int \sin^{-1} u\,(2u\,du)=2\int u\sin^{-1} u\,du.
$$

We now evaluate

$$
I= \int u\sin^{-1} u\,du
$$

using integration by parts. Let

$$
\begin{aligned}
v &= \sin^{-1} u \quad\Longrightarrow\quad dv =\frac{1}{\sqrt{1-u^2}}\,du,\[1mm]
dw &= u\,du \quad\Longrightarrow\quad w = \frac{u^2}{2}.
\end{aligned}
$$

Then

$$
I = v\,w - \int w\,dv = \frac{u^2}{2}\sin^{-1} u - \int \frac{u^2}{2}\frac{du}{\sqrt{1-u^2}}.
$$

So

$$
I = \frac{u^2}{2}\sin^{-1} u - \frac{1}{2}\int \frac{u^2}{\sqrt{1-u^2}}\,du.
$$

Returning to our original integral,

$$
\int \sin^{-1}\sqrt{x}\,dx = 2 I = u^2\sin^{-1}u - \int \frac{u^2}{\sqrt{1-u^2}}\,du.
$$

Next, we simplify the remaining integral. Notice that

$$
u^2 = 1 - (1-u^2).
$$

Thus we can write

$$
\int \frac{u^2}{\sqrt{1-u^2}}\,du = \int \frac{1}{\sqrt{1-u^2}}\,du - \int \frac{1-u^2}{\sqrt{1-u^2}}\,du.
$$

The two integrals on the right are standard:

1. $$\displaystyle \int \frac{du}{\sqrt{1-u^2}}=\sin^{-1} u$$.

2. $$\displaystyle \int \frac{1-u^2}{\sqrt{1-u^2}}\,du = \int \sqrt{1-u^2}\,du$$.

A standard result is

$$
\int \sqrt{1-u^2}\,du = \frac{1}{2}\left(u\sqrt{1-u^2}+\sin^{-1} u\right).
$$

Thus,

$$
\int \frac{u^2}{\sqrt{1-u^2}}\,du = \sin^{-1} u - \frac{1}{2}\left(u\sqrt{1-u^2}+\sin^{-1} u\right).
$$

This simplifies to

$$
\int \frac{u^2}{\sqrt{1-u^2}}\,du = \frac{1}{2}\sin^{-1} u - \frac{1}{2}u\sqrt{1-u^2}.
$$

Returning to our integral we now have

$$
\int \sin^{-1}\sqrt{x}\,dx = u^2\sin^{-1}u - \left[\frac{1}{2}\sin^{-1} u - \frac{1}{2}u\sqrt{1-u^2}\right].
$$

That is,

$$
\int \sin^{-1}\sqrt{x}\,dx = u^2\sin^{-1}u - \frac{1}{2}\sin^{-1} u + \frac{1}{2}u\sqrt{1-u^2}+C.
$$

Recall that $$ u = \sqrt{x} $$ so that

$$
u^2 = x \quad \text{and} \quad u\sqrt{1-u^2} = \sqrt{x}\sqrt{1-x}=\sqrt{x-x^2}.
$$

Thus, the integral becomes

$$
\int \sin^{-1}\sqrt{x}\,dx = \left(x-\frac{1}{2}\right)\sin^{-1}\sqrt{x}+\frac{1}{2}\sqrt{x-x^2}+C.
$$

This result corresponds to option A:

$$
\left(x-\frac{1}{2}\right) \sin^{-1} \sqrt{x}+\frac{1}{2} \sqrt{x-x^{2}}+C.
$$
The integral evaluates to $$ \left(x-\frac{1}{2}\right) \sin^{-1} \sqrt{x}+\frac{1}{2} \sqrt{x-x^{2}}+C $$, which corresponds to option A.

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