Evaluate the indefinite integral.∫esecθsecθtanθdθ\int_{}^{}e^{\sec\theta}\sec\theta\tan\theta d\theta

e sec ⁡ θ + C e^{\sec\theta}+C

Evaluate the indefinite integral. ∫ e sec θ sec θ tan θ d θ \int_{}^{}e^{\sec\theta}\sec\theta\tan\theta d\theta

we're asked to evaluate the indefinite integral of e to the power of secant theta times secant theta tangent theta d theta. Now on first glance, this might look like a rather challenging integral, but let's go ahead and go back to what we know. We know that if we see some function in that power of e, that's typically what we should choose as u when making a substitution. So let's see what happens when we do that. If u is equal to secant of theta, then du is equal to the derivative of that. So secant theta tangent theta times d theta. And looking back at our integral, that's exactly what else is left over in my integral. So if that secant theta is equal to U, then DU has nicely appeared appeared in my integral here, and I am able to rewrite this as e to the u du. Now from our rule, we know that this just results in e to the u plus c. And again, since we know what that u is, we can plug that back in. So that is the secant of theta. And this is my final answer, e to the power of secant theta plus my constant of integration, c. See you in the next one.

Evaluate the indefinite integral. ∫ e sec ⁡ θ sec ⁡ θ tan ⁡ θ d θ \int_{}^{}e^{\sec\theta}\sec\theta\tan\theta d\theta

e sec ⁡ θ + C e^{\sec\theta}+C

− e sec ⁡ θ c o s 2 ⁡ θ + C -\frac{e^{\sec\theta}}{cos^2⁡\theta}+C

− e sec ⁡ θ + C -e^{\sec\theta}+C

e sec ⁡ θ c o s 2 ⁡ θ + C \frac{e^{\sec\theta}}{cos^2⁡\theta}+C

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0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
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10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals of Exponential Functions

Calculus

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals of Exponential Functions

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Multiple Choice

Evaluate the indefinite integral.
$\int_{}^{} e^{\sec θ} \sec θ \tan θ d θ$

$- \frac{e^{\sec θ}}{c o s^{2} θ} + C$

$e^{\sec θ} + C$

$- e^{\sec θ} + C$

$\frac{e^{\sec θ}}{c o s^{2} θ} + C$

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Verified step by step guidance

Step 1: Recognize the structure of the integral. The given integral is ∫e^{secθ} secθ tanθ dθ. Notice that the integrand contains e^{secθ}, secθ, and tanθ, which suggests a substitution method might simplify the problem.

Step 2: Choose an appropriate substitution. Let u = secθ. Then, the derivative of secθ is d(secθ)/dθ = secθ tanθ. This means that du = secθ tanθ dθ. Substituting this into the integral simplifies it significantly.

Step 3: Rewrite the integral in terms of u. Using the substitution u = secθ and du = secθ tanθ dθ, the integral becomes ∫e^u du. This is a standard exponential integral.

Step 4: Solve the simplified integral. The integral of e^u with respect to u is simply e^u + C, where C is the constant of integration.

Step 5: Back-substitute to return to the original variable. Since u = secθ, replace u in the result with secθ. The final expression for the indefinite integral is e^{secθ} + C.