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)=5tan⁡θ+2cot⁡θ+Cg\left(\theta\right)=5\tan\theta+2\cot\theta+Cg(θ)=5tanθ+2cotθ+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 if the integral of five times the secant squared of theta is going to be minus the integral of two 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 five, and I'll take this two 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 five times the integral of secant squared theta, and then we're going to have minus two 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 five times the tangent of theta, and then we're going to have this whole thing minus this integral. Now we're going to have two 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 five, and then that's gonna be tangent theta is going to be plus two 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)=∫(5sec⁡2θ−2csc⁡2θ)dθg\left(\theta\right)=\int^{}\left(5\$$sec^2$$\theta-2\$$csc^2$$\theta\right)d\thetag(θ)=∫(5sec2θ−2csc2θ)dθ

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

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

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

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

12. Trigonometric Functions

Integrals of Basic Trig Functions

Business Calculus

12. Trigonometric Functions

Integrals of Basic Trig Functions

Struggling with Business Calculus?

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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

Step 1: Recognize that the problem involves finding the indefinite integral of the given function g(θ) = ∫(5sec²θ - 2csc²θ)dθ. This means we need to integrate each term in the expression separately.

Step 2: Recall the standard integral formulas: ∫sec²θ dθ = tanθ + C and ∫csc²θ dθ = -cotθ + C. These will be used to evaluate the integral of each term.

Step 3: Apply the integral formula for sec²θ to the first term, 5sec²θ. The integral of 5sec²θ is 5∫sec²θ dθ = 5tanθ.

Step 4: Apply the integral formula for csc²θ to the second term, -2csc²θ. The integral of -2csc²θ is -2∫csc²θ dθ = -2(-cotθ) = 2cotθ.

Step 5: Combine the results from Steps 3 and 4, and include the constant of integration C. The final expression for g(θ) is g(θ) = 5tanθ + 2cotθ + C.

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Struggling with Business Calculus?

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\(\theta\)g(θ)=∫(5sec2θ−2csc2θ)dθ

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Integrals of Basic Trig Functions practice set

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\(\theta\)$