Find g(x)g\left(x\right)g(x) by evaluating the following indefinite integral.g(x)=∫(sin2x−100cscxcotx+cos2x)dxg\left(x\right)=\int\left(\sin^2x-100\csc x\cot x+\cos^2x\right)dxg(x)=∫(sin2x−100cscxcotx+cos2x)dx

g ( x ) = 100 csc ⁡ x + x + C g\left(x\right)=100\csc x+x+C g ( x ) = 100 csc x + x + C

Find g ( x ) g\left(x\right) g ( x ) by evaluating the following indefinite integral. g ( x ) = ∫ ( sin 2 x − 100 csc x cot x + cos 2 x ) d x g\left(x\right)=\int\left(\sin^2x-100\csc x\cot x+\cos^2x\right)dx g ( x ) = ∫ ( sin 2 x − 100 csc x cot x + cos 2 x ) d x

Let's see if we can go about solving this problem. So here we're asked to evaluate the following indefinite integral and solve for this function g of x. Now right off the bat, this situation looks like a headache, because we have some complicated trigonometric functions in here. We're also dealing with the sine and cosine terms being squared. There's just a lot of weird stuff going on. But right off the bat, there's something you can actually do to make your life a whole lot easier when solving this problem, and that's recognize that there's a sine squared in this problem and a cosine squared in this problem. Recall that sine squared plus cosine squared equals 1, based on the trigonometric identities that we've seen. So the pythagorean identity says that sine squared x plus cosine squared x is always gonna be equal to 1. So what that means is I can actually rewrite this integral by just rearranging everything that we have in these parentheses. So we're going to have negative 100 cosecant x cotangent x, that's still gonna be by itself. So I have negative 100 cosecant x, and it's gonna be times cotangent x. But then we're going to have this, and that's going to be plus, and then we'll have sine squared x, and we're going to have plus cosine squared x, and we're integrating this whole thing with respect to x. Now this entire piece right here, based on the Pythagorean identity, is equal to 1. So what that means is that this is going to simplify to the indefinite integral of negative 100 cosecant x cotangent x, and it's going to be plus. And then this whole thing will just go to 1 because it's equal to 1 right here. So that's what it's gonna become. And so we'll have this plus 1, and then that is going to be our integral. So notice how it's a lot more simple to solve now that we have this. Now from here, I can apply the rules for integrals, the sum and difference rule mainly. So we're going to have the integral of negative 100 times cosecant x cotangent x, and we're going to have plus, and we're gonna have the integral of 1 with respect to x. Now at this point, I can also use the constant rule, which is to take this negative a 100 out of side the integral and leave just the trig functions in here. So we're gonna have negative 100 as going to be the integral of cosecant x, cotangent x. And keep in mind, this is all with respect to x too. I need to write that up here as well. So we won't have the integral here of 1 with respect to x, so plus the integral of 1 with with respect to x. You could pull it outside the integral too, but since it's just one, I'll just leave it inside the integral there. And then keep in mind, we'd have a d x up here as well. So we're doing this all with respect to x. And then from here, I need to evaluate the integrals that I see. Now the integral of cosecant, cotangent with respect to x, this is actually something we've seen. This comes up to negative cosecant. So what that means, we have negative 100, and it's gonna be times negative cosecant of x. Right? And then we're gonna have plus in the integral of 1 with with respect to x. That's just gonna be x. And then we have plus the constant of integration c. I can cancel the negative sign of the 100 and the negative cosecant. So this is gonna leave me with 100 cosecant x plus x plus c as my solution for g of x. So this right here would be the solution to the problem, and that's how you can solve it. So So as you can see, it's pretty straightforward as long as you have, and you're familiar with some of the common integrals that you see for trig functions, we applied that right here. But for this case, we just use the Pythagorean identity to simplify this even further, giving us an integral that we can actually solve using the rules that we already know about for integration. So hope you found this video helpful, and let's move on.

Find g ( x ) g\left(x\right) g ( x ) by evaluating the following indefinite integral. g ( x ) = ∫ ( sin ⁡ 2 x − 100 csc ⁡ x cot ⁡ x + cos ⁡ 2 x ) d x g\left(x\right)=\int\left(\sin^2x-100\csc x\cot x+\cos^2x\right)dx g ( x ) = ∫ ( sin 2 x − 100 csc x cot x + cos 2 x ) d x

g ( x ) = 100 csc ⁡ x + x + C g\left(x\right)=100\csc x+x+C g ( x ) = 100 csc x + x + C

g ( x ) = − 100 csc ⁡ x + x + C g\left(x\right)=-100\csc x+x+C g ( x ) = − 100 csc x + x + C

g ( x ) = 100 csc ⁡ x + C g\left(x\right)=100\csc x+C g ( x ) = 100 csc x + C

g ( x ) = − 100 csc ⁡ x + C g\left(x\right)=-100\csc x+C g ( x ) = − 100 csc x + C

7. Antiderivatives & Indefinite Integrals

Integrals of Trig Functions

Calculus

7. Antiderivatives & Indefinite Integrals

Integrals of Trig Functions

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Multiple Choice

Find $g\left(x\right)$ by evaluating the following indefinite integral.
$g\left(x\right)=\int\left(\sin^2x-100\csc x\cot x+\cos^2x\right)dx$

$g\left(x\right)=100\csc x+x+C$

$g\left(x\right)=-100\csc x+x+C$

$g\left(x\right)=100\csc x+C$

$g\left(x\right)=-100\csc x+C$

Verified step by step guidance

Identify the integral to be evaluated: $ g(x) = \int (\sin^2 x - 100 \csc x \cot x + \cos^2 x) \, dx $.

Recognize that $ \sin^2 x + \cos^2 x = 1 $ is a trigonometric identity, which simplifies the integral to $ \int (1 - 100 \csc x \cot x) \, dx $.

Separate the integral into two parts: $ \int 1 \, dx - \int 100 \csc x \cot x \, dx $.

Evaluate the first integral: $ \int 1 \, dx = x $.

Evaluate the second integral: $ \int 100 \csc x \cot x \, dx = -100 \csc x $, using the fact that the derivative of $ \csc x $ is $ -\csc x \cot x $. Combine the results and add the constant of integration $ C $.

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