Evaluate the integral.∫(21−x2−13x)dx\int_{}^{}\left(\frac{2}{\sqrt{1-x^2}}-\frac{1}{3\sqrt{x}}\right)dx

2 s i n − 1 ⁡ x − 2 3 x + C 2sin^{-1}⁡x-\frac23\sqrt{x}+C

Evaluate the integral. ∫ ( 2 1 − x 2 − 1 3 x ) d x \int_{}^{}\left(\frac{2}{\sqrt{1-x^2}}-\frac{1}{3\sqrt{x}}\right)dx

Here we're asked to evaluate the integral of two over the square root of one minuteus x squared minus one over three root x dx. Now looking at the two terms in my integrand here, this first one, two over the square root of one minuteus x squared, looks really similar to this here that results in an inverse sine, but it just has a two in that numerator instead of a one, and that's okay because it just varies by a constant. This just means that my antiderivative here is gonna be two times the inverse sine of x. Then looking at this second term here, one over three root x. Remember, just because you have a square root x in the denominator does not automatically mean we're going to be dealing with some sort of inverse trig functions. As we continue to learn more and more rules for integration, we can't forget the rules that we already know. We can just use a general power rule here. This is the same thing as a minus one third x to the power of a negative one half. We're just going to use a general power rule here, adding one to that power and then dividing by that new power. This ultimately gives us here minus two thirds x to the power of positive one half. So we've just combined those constants here, one third with that two, and then we're gonna add on that constant of integration c there. Now I can rewrite this x to the power of one half as just the square root of x, and this gives me my final answer here for this integral. Let us know if you have any questions, and I'll see you in the next one.

Evaluate the integral. ∫ ( 2 1 − x 2 − 1 3 x ) d x \int_{}^{}\left(\frac{2}{\sqrt{1-x^2}}-\frac{1}{3\sqrt{x}}\right)dx

2 s i n − 1 ⁡ x − 2 3 x + C 2sin^{-1}⁡x-\frac23\sqrt{x}+C

2 c o s − 1 ⁡ x − 2 3 x + C 2cos^{-1}⁡x-\frac23\sqrt{x}+C

c o s − 1 ⁡ x − 2 3 x + C cos^{-1}⁡x-\frac23\sqrt{x}+C

s i n − 1 x − 2 3 x + C sin^{-1}x-\frac23\sqrt{x}+C

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

Evaluate the integral.
$\int_{}^{} (\frac{2}{\sqrt{1 - x^{2}}} - \frac{1}{3 \sqrt{x}}) d x$

$2 c o s^{- 1} x - \frac{2}{3} \sqrt{x} + C$

$2 s i n^{- 1} x - \frac{2}{3} \sqrt{x} + C$

$c o s^{- 1} x - \frac{2}{3} \sqrt{x} + C$

$s i n^{- 1} x - \frac{2}{3} \sqrt{x} + C$

Verified step by step guidance

Step 1: Break the integral into two separate terms for easier evaluation. The given integral is ∫(2/√(1-x²) - 1/(3√x)) dx. Rewrite it as two separate integrals: ∫(2/√(1-x²)) dx - ∫(1/(3√x)) dx.

Step 2: Recognize the first term, ∫(2/√(1-x²)) dx, as a standard integral form. The integral of 1/√(1-x²) is sin⁻¹(x), so the integral of 2/√(1-x²) is 2sin⁻¹(x).

Step 3: For the second term, ∫(1/(3√x)) dx, rewrite 1/(3√x) as (1/3)x^(-1/2). Use the power rule for integration, which states ∫xⁿ dx = (xⁿ⁺¹)/(n+1) + C, where n ≠ -1. Here, n = -1/2, so the integral becomes (1/3) * (x^(1/2)/(1/2)) = (2/3)√x.

Step 4: Combine the results of the two integrals. The first term contributes 2sin⁻¹(x), and the second term contributes -(2/3)√x (note the subtraction from the original integral).

Step 5: Add the constant of integration, C, to account for the indefinite integral. The final expression is 2sin⁻¹(x) - (2/3)√x + C.

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