Evaluate the definite integral in terms of an inverse trig function. ∫ 0 3 3 d x 4 − 9 x 2 \int_0^{\frac{\sqrt3}{3}}\frac{dx}{\sqrt{4-9x^2}} ∫ 0 3 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"> 4 − 9 x 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"> d x

Here we want to evaluate the definite integral in terms of an inverse trig function and we're looking at the definite integral from zero to root three over three of DX over the square root of four minus nine X squared. Now Now I'm going to start by ignoring my bounds and then I'm going to address them at the end just so that I can really focus on my antiderivative before I get to those bounds. So I'm going to rewrite this integral slightly, treating it as an indefinite integral for now, of DX over the square root of two squared as opposed to four minus three x squared. So just rewriting them so that we can clearly see what U and A are going to be. Now looking at this here, I can see that it takes the form of one over the square root of A squared minus U squared, specifically with an a value here of two and a u of three x. Now if u is equal to three x, that means that du has to be equal to three d x. Now I don't have that three currently, so I need to multiply by it. And if I'm multiplying by three, I also need to multiply by its reciprocal one third so that I'm not actually changing the value of my integral. So now I have a u and I have this du completing my substitution, And I can rewrite this as one third times the integral of one over the square root of a squared minus u squared du, which is exactly my formula here. So I know that this will result in one third times the inverse sine of u over a plus c, even though I know I'm not going to have to worry about that plus c in just a moment. So I'm going to go ahead and plug my u and a values back in. Remember, u is three x a is two. So this is one third times the inverse sine of three x over two and then plus c. Now again, here we don't really have to worry about that plus c and now I can go ahead and stick my bounds on there. This is bounded from zero to root three over three since we have this back in terms of our original variable x. And now all I need to do is plug those bounds in. So keeping that one third on the outside here, just so I don't have to write it twice, this is one third times the inverse sign of three times or root three over three over two minus the inverse sine of three times zero over two, which is just zero. Now looking at this here, these threes are going to cancel. So I'm really looking at here one third times the inverse sine of root three over two minus the inverse sine of zero. Now the inverse sine of zero is zero. So this term goes away and the inverse sine of root three over two. Remember, we're thinking about what angle we're looking at that makes the sine root three over two, and that is pi over three. So this is one third times pi over three. Multiplying across here gives me a final answer of pi over pi over nine for this definite integral. Feel free to let us know if you have any questions in the comments. I'll see you in the next one.

Evaluate the definite integral in terms of an inverse trig function. ∫ 0 3 3 d x 4 − 9 x 2 \int_0^{\frac{\sqrt3}{3}}\frac{dx}{\sqrt{4-9x^2}} ∫ 0 3 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> 4 − 9 x 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> d x

− π 3 -\frac{\pi}{3} − 3 π

− π 9 -\frac{\pi}{9} − 9 π

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 definite integral in terms of an inverse trig function.
$$\int$_0^{$\frac{\sqrt3}{3}$}$\frac{dx}{\sqrt{4-9x^2}$}$

$\frac{\pi}{3}$

$-$\frac{\pi}{3}$$

$pi over 9$

$-$\frac{\pi}{9}$$

Verified step by step guidance

Step 1: Recognize the integral's form. The given integral resembles the standard form of an inverse trigonometric function: ∫ dx / √(a² - u²) = arcsin(u/a) + C. Here, a² corresponds to 4, and u² corresponds to 9x².

Step 2: Rewrite the integral to match the standard form. Factor out constants to make the denominator look like √(a² - u²). Specifically, rewrite 4 - 9x² as (2² - (3x)²).

Step 3: Perform a substitution to simplify the integral. Let u = 3x, so that du = 3 dx, or dx = du / 3. Update the limits of integration accordingly: when x = 0, u = 0, and when x = √3/3, u = 1.

Step 4: Substitute into the integral. Replace x, dx, and the limits with the new variable u. The integral becomes (1/3) ∫ du / √(4 - u²), which simplifies to (1/3) ∫ du / √(2² - u²).

Step 5: Apply the arcsin formula. Using the standard result ∫ du / √(a² - u²) = arcsin(u/a), the integral evaluates to (1/3) * arcsin(u/2). Substitute back the limits of integration (u = 0 to u = 1) to find the final result.

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Evaluate the definite integral in terms of an inverse trig function.∫033dx4−9x2\(\int\)_0^{\(\frac{\sqrt3}{3}\)}\(\frac{dx}{\sqrt{4-9x^2}\)}∫0334−9x2dx

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Evaluate the definite integral in terms of an inverse trig function.
$\(\int\)_0^{\(\frac{\sqrt3}{3}\)}\(\frac{dx}{\sqrt{4-9x^2}\)}$