Find g(θ)g\left(\theta\right)g(θ) by evaluating the following indefinite integral.g(θ)=∫(5sec2θ−2csc2θ)dθg\left(\theta\right)=\int^{}\left(5\sec^2\theta-2\csc^2\theta\right)d\thetag(θ)=∫(5sec2θ−2csc2θ)dθ

g ( θ ) = 5 tan ⁡ θ + 2 cot ⁡ θ + C g\left(\theta\right)=5\tan\theta+2\cot\theta+C g ( θ ) = 5 tan θ + 2 cot θ + C

Find g ( θ ) g\left(\theta\right) g ( θ ) by evaluating the following indefinite integral. g ( θ ) = ∫ ( 5 sec 2 θ − 2 csc 2 θ ) d θ g\left(\theta\right)=\int^{}\left(5\sec^2\theta-2\csc^2\theta\right)d\theta g ( θ ) = ∫ ( 5 sec 2 θ − 2 csc 2 θ ) d θ

this problem, we need to solve for the function g of theta by evaluating the integral that we see right here. Now what I'm going to do is apply the rules that I know for integrals. So one of the things I can see is that we have this minus sign between these two terms. So So I can integrate these terms separately. So the integral of 5 times the secant squared of theta is going to be minus the integral of 2 cosecant squared theta. Now at this point, I'm going to take any constants that I see and pull them outside of the integrals. Also, keep in mind that notice we're integrating with respect to theta here. So we're gonna have a d theta on the end of both of these integrals. Now what I'll do is I'll take this 5, and I'll take this 2 since these are constants attached to these trig functions that we see, and I'll pull them outside the integrals to make things more simple. So we're going to have 5 times the integral of secant squared theta, and then we're going to have minus 2 times the integral of cosecant squared theta. And this is what we get. Now from here, I need to apply what I know for integrating trig functions like this. Now we talked about what the integral of secant squared theta comes out to tangent theta. So what that means is this whole term here is going to be tangent theta, so we're going to have 5 times the tangent of theta, and then we're going to have this whole thing minus this integral. Now we're going to have 2 times the integral of cosecant squared theta. And we talked about the integral of cosecant squared theta comes up to negative cotangent theta. So we get negative cotangent theta right there. And then don't forget we need to add the plus c for the cost of integration at the end of any indefinite integral. Now we can clean this up a little bit because this minus sign here is gonna cancel with the negative sign right there. So what we're gonna end up with is 5, and then that's gonna be tangent theta is going to be plus 2 cotangent theta plus the constant of integration c. That is our function g of theta and the solution to this problem. So this is how you can solve these indefinite integrals where you are dealing with trig functions and some of these more complicated trig functions, like when we see the secants or cosecants. Just think about the rules that we've learned about for integrating trig functions and apply them. But it's usually best to apply the other rules that we've learned as well, like the constant and the multiple rules and the sum and difference rules, and to go ahead and get these trig functions by themselves so it's really straightforward to integrate. Hope you found this video helpful. Let's move on.

Find g ( θ ) g\left(\theta\right) g ( θ ) by evaluating the following indefinite integral. g ( θ ) = ∫ ( 5 sec ⁡ 2 θ − 2 csc ⁡ 2 θ ) d θ g\left(\theta\right)=\int^{}\left(5\sec^2\theta-2\csc^2\theta\right)d\theta g ( θ ) = ∫ ( 5 sec 2 θ − 2 csc 2 θ ) d θ

g ( θ ) = 5 tan ⁡ θ + 2 cot ⁡ θ + C g\left(\theta\right)=5\tan\theta+2\cot\theta+C g ( θ ) = 5 tan θ + 2 cot θ + C

g ( θ ) = 5 tan ⁡ θ − 2 cot ⁡ θ + C g\left(\theta\right)=5\tan\theta-2\cot\theta+C g ( θ ) = 5 tan θ − 2 cot θ + C

g ( θ ) = 5 cot ⁡ θ + 2 tan ⁡ θ + C g\left(\theta\right)=5\cot\theta+2\tan\theta+C g ( θ ) = 5 cot θ + 2 tan θ + C

g ( θ ) = 5 cot ⁡ θ − 2 tan ⁡ θ + C g\left(\theta\right)=5\cot\theta-2\tan\theta+C g ( θ ) = 5 cot θ − 2 tan θ + C

7. Antiderivatives & Indefinite Integrals

Integrals of Trig Functions

Calculus

7. Antiderivatives & Indefinite Integrals

Integrals of Trig Functions

Struggling with Calculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Find $g\left(\theta\right)$ by evaluating the following indefinite integral.
$g\left(\theta\right)=\int^{}\left(5\sec^2\theta-2\csc^2\theta\right)d\theta$

$g\left(\theta\right)=5\tan\theta+2\cot\theta+C$

$g\left(\theta\right)=5\tan\theta-2\cot\theta+C$

$g\left(\theta\right)=5\cot\theta+2\tan\theta+C$

$g\left(\theta\right)=5\cot\theta-2\tan\theta+C$

Verified step by step guidance

Start by recognizing the integral: $ g(\theta) = \int (5 \sec^2 \theta - 2 \csc^2 \theta) \, d\theta $. This is an indefinite integral, meaning we are looking for a function whose derivative is the integrand.

Recall the derivatives of the trigonometric functions: $ \frac{d}{d\theta} \tan \theta = \sec^2 \theta $ and $ \frac{d}{d\theta} \cot \theta = -\csc^2 \theta $. These will help us identify the antiderivatives.

Break down the integral into two separate parts: $ \int 5 \sec^2 \theta \, d\theta $ and $ \int -2 \csc^2 \theta \, d\theta $. This allows us to integrate each term individually.

For the first part, $ \int 5 \sec^2 \theta \, d\theta $, use the fact that the antiderivative of $ \sec^2 \theta $ is $ \tan \theta $. Therefore, $ \int 5 \sec^2 \theta \, d\theta = 5 \tan \theta + C_1 $.

For the second part, $ \int -2 \csc^2 \theta \, d\theta $, use the fact that the antiderivative of $ -\csc^2 \theta $ is $ \cot \theta $. Therefore, $ \int -2 \csc^2 \theta \, d\theta = 2 \cot \theta + C_2 $. Combine these results to get $ g(\theta) = 5 \tan \theta + 2 \cot \theta + C $, where $ C $ is the constant of integration.

5:28m

Watch next

Master Integrals Resulting in Basic Trig Functions with a bite sized video explanation from Patrick

Start learning

Integrals of Trig Functions practice set

Problem sets built by lead tutors

Expert video explanations

Go to Practice