Evaluate the indefinite integral.∫θ∙sec2(5θ2+1)dθ\int_{}^{}\theta\bullet sec^2\left(5\theta^2+1\right)d\theta∫θ∙sec2(5θ2+1)dθ

1 10 tan ⁡ ( 5 θ 2 + 1 ) + C \frac{1}{10}\tan\left(5\theta^2+1\right)+C 10 1 tan ( 5 θ 2 + 1 ) + C

Evaluate the indefinite integral. ∫ θ ∙ s e c 2 ( 5 θ 2 + 1 ) d θ \int_{}^{}\theta\bullet sec^2\left(5\theta^2+1\right)d\theta ∫ θ ∙ se c 2 ( 5 θ 2 + 1 ) d θ

this problem, we want to evaluate the indefinite integral of theta times the secant squared of five theta squared plus one d theta. Now even though our variable here is the theta, we can still use substitution the same way the variable that we're working with in our original integral does not ultimately matter. So anytime we see an x in our steps here, we can really just consider that to be theta just because that's what we're given in our integrand. So let's go ahead and get started with our steps for substitution here in order to evaluate this integral. The first thing that we wanna do is choose u and find du. Now because here we're working with a trig function, that can often kind of trip people up as they decide what u should be. Now looking at this particular integral, we can see that we have this five theta squared plus one that we're taking the secant squared of. So that is inside the argument of our trig function. Now because u is often going to be some inside function, it makes sense that that would be a good choice for you. So if I make u equal to five theta squared plus one, what does that make du? It's going to be the derivative of that five theta squared plus one times d theta. Now the derivative here is going to be 10 theta and then this is multiplied by d theta. So u is five theta squared plus one, that makes DU 10 theta d theta. Now we can move on to step two, rewriting our integral only in terms of this u and du. Now looking at my integral here, I have this u but since du is 10 theta d theta, I have theta d theta in my integral, but I do not have 10. So that means that I wanna go ahead and multiply it by this missing constant and its reciprocal in order to make this substitution work. So that means that ultimately in this integral, I am multiplying by one tenth and 10, a the constant that I want and its reciprocal. So I'm gonna go ahead and rewrite this in an order that makes a little bit more sense here, keeping that one tenth constant on the outside of my integral. This is one tenth times the secant squared of five theta squared plus one and then times 10 theta d theta. I'm just rearranging everything that I have over here. Now we can clearly see that we have the secant squared of u and then times du. So rewriting this in terms of that u and du, this is one tenth times the integral of secant squared of u and then times du. So we can go ahead and continue on with step three here, integrating with respect to u. So we wanna think about our antiderivative here. What is the antiderivative of secant squared? It is the tangent. So this is going to give us a one tenth times the tangent of u plus that constant of integration c since this is an indefinite integral. So we have taken that integral and we can now replace u with our original function here of theta. So this was five theta squared plus one. That's what we're replacing this u with. So this is going to be one tenth times the tangent of five theta squared plus one plus that constant of integration c. And this is our final answer. Let us know if you have any questions. I'll see you in the next one.

Evaluate the indefinite integral. ∫ θ ∙ s e c 2 ( 5 θ 2 + 1 ) d θ \int_{}^{}\theta\bullet sec^2\left(5\theta^2+1\right)d\theta ∫ θ ∙ se c 2 ( 5 θ 2 + 1 ) d θ

1 10 tan ⁡ ( 5 θ 2 + 1 ) + C \frac{1}{10}\tan\left(5\theta^2+1\right)+C 10 1 tan ( 5 θ 2 + 1 ) + C

10 tan ⁡ ( 5 θ 2 + 1 ) + C 10\tan\left(5\theta^2+1\right)+C 10 tan ( 5 θ 2 + 1 ) + C

10 s e c 2 ( 5 θ + 1 ) + C 10sec^2\left(5\theta+1\right)+C 10 se c 2 ( 5 θ + 1 ) + C

1 20 θ 2 tan ⁡ ( 5 θ 2 + 1 ) + C \frac{1}{20}\theta^2\tan\left(5\theta^2+1\right)+C 20 1 θ 2 tan ( 5 θ 2 + 1 ) + C

12. Trigonometric Functions

Integrals of Basic Trig Functions

Business Calculus

12. Trigonometric Functions

Integrals of Basic Trig Functions

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Multiple Choice

Evaluate the indefinite integral.
$\int_{}^{}\theta\bullet sec^2\left(5\theta^2+1\right)d\theta$

$10\tan\left(5\theta^2+1\right)+C$

$\frac{1}{10}\tan\left(5\theta^2+1\right)+C$

$10sec^2\left(5\theta+1\right)+C$

$\frac{1}{20}\theta^2\tan\left(5\theta^2+1\right)+C$

Verified step by step guidance

Step 1: Recognize that the integral involves a product of θ and sec²(5θ² + 1). This suggests that substitution might be a useful technique. Let u = 5θ² + 1, which simplifies the argument of the sec² function.

Step 2: Differentiate u with respect to θ to find du. Since u = 5θ² + 1, we have du/dθ = 10θ. Therefore, du = 10θ dθ.

Step 3: Rewrite the integral in terms of u. Substitute u = 5θ² + 1 and du = 10θ dθ into the integral. This transforms the integral into (1/10) ∫ sec²(u) du.

Step 4: Recall the integral of sec²(u) with respect to u. The integral of sec²(u) is tan(u) + C, where C is the constant of integration.

Step 5: Substitute back u = 5θ² + 1 into the result to return to the original variable. The final expression becomes (1/10) tan(5θ² + 1) + C.

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