Find the indefinite integral.∫arcsin4x1−x2dx\int_{}^{}\frac{arcsin^4x}{\sqrt{1-x^2}}dx

1 5 a r c s i n 5 x + C \frac15arcsin^5x+C

Find the indefinite integral. ∫ a r c s i n 4 x 1 − x 2 d x \int_{}^{}\frac{arcsin^4x}{\sqrt{1-x^2}}dx

we're asked to find the indefinite integral of arcsine to the power of four of x over the square root of one minuteus x squared DX. Now it's important to remember here that arcsine and inverse sine are the exact same thing here. Now when working with this problem, I can tell that I need to do a substitution. And oftentimes when making a substitution, we want to choose some sort of inside function, whether that's under a radical, in parentheses, or raised to a power. Now in this case, I do see a function under a radical, but making that equal to u won't really help me here. So instead, I want to look at my arcsine function, which is being raised to the power of four. What if I set u equal to the arcsine of x? If I set u equal to the arcsine of x, du is equal to the derivative of that, which is a one over the square root of one minus x squared times d x. Now that's exactly what I see show up in my integral here. So this is a good substitution. Now having made this arcsine u and having a du there, I can then rewrite this as the integral of u to the power of four du. So this just becomes a basic power rule integral. So applying my power rule here, this becomes one fifth u to the power of five plus c, and I can plug u back in here. I know it's the arc sine of x. So this is one fifth times the arc sine. I'm going to go ahead and put that power there in between my arc sign and my X and then add on that C there. And this gives me my final answer. Let us know if you have any questions and let's keep going.

Find the indefinite integral. ∫ a r c s i n 4 x 1 − x 2 d x \int_{}^{}\frac{arcsin^4x}{\sqrt{1-x^2}}dx

1 5 a r c s i n 5 x + C \frac15arcsin^5x+C

a r c s i n 5 x + C arcsin^5x+C

1 3 a r c s i n 3 x + C \frac13arcsin^3x+C

− 1 5 a r c c o s 5 x + C -\frac15arccos^5x+C

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0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
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5. Graphical Applications of Derivatives6h 2m
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7. Antiderivatives & Indefinite Integrals1h 26m
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10. Physics Applications of Integrals 3h 16m
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11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
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- Introduction to Power Series
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16. Parametric Equations & Polar Coordinates7h 58m

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Inverse Trigonometric Functions

Calculus

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Inverse Trigonometric Functions

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Multiple Choice

Find the indefinite integral.
$\int_{}^{} \frac{a r c s i n^{4} x}{\sqrt{1 - x^{2}}} d x$

$a r c s i n^{5} x + C$

$\frac{1}{3} a r c s i n^{3} x + C$

$- \frac{1}{5} a r c c o s^{5} x + C$

$\frac{1}{5} a r c s i n^{5} x + C$

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Verified step by step guidance

Step 1: Recognize the structure of the integral. The given integral is ∫(arcsin^4(x) / √(1 - x^2)) dx. Notice that the derivative of arcsin(x) is 1 / √(1 - x^2), which suggests that substitution involving arcsin(x) might simplify the problem.

Step 2: Perform substitution. Let u = arcsin(x). Then, du = 1 / √(1 - x^2) dx. This substitution transforms the integral into ∫u^4 du.

Step 3: Integrate the new expression. The integral ∫u^4 du is a standard power rule integral. Use the formula ∫u^n du = u^(n+1) / (n+1) + C, where n ≠ -1. Here, n = 4, so the integral becomes u^5 / 5 + C.

Step 4: Substitute back the original variable. Recall that u = arcsin(x). Replace u in the result to get (arcsin(x))^5 / 5 + C.

Step 5: Simplify the final expression. The final answer is (1/5) * (arcsin(x))^5 + C, which matches the correct answer provided in the problem.