Evaluate the integrals in Exercises 37–44.
∫ eᵗ √[tan²(e...

Question

Evaluate the integrals in Exercises 37–44.

∫ eᵗ √[tan²(eᵗ) + 1] dt

Accepted Answer

Welcome back everyone. Assuming 2 of 3 to the power of t is greater than or equal to 0, evaluate the following integral. Integral of 3 to the power of t multiplied by square root of 1 plus tangent squared of 3 to the power of t dt. Let's begin this problem by simplifying the expression within the integral. Specifically, what we can do is notice that we have 1 + tangent squared of 3 to the power of t. Using trigonometric identities, we can essentially say that this is equal to square of 3 to the power of t. And we're taking square root of. C and squared of 3s the power of t. Which is the absolute value of 2 of 3 to the power of t. And we know that C of 3 to the power of t is greater than or equal to 0, so we can drop the absolute value. So now we are ready to rewrite the integral as the integral of 3 to the power of t. See Of 3 to the power of t dt. And what we're going to do is apply substitution. We're going to set you equal to 3 to the power of t and remember that the next step is to find the derivative of you with respect to t. Which is the derivative of 3 to the power of t, and it is 3 to the power of tl and of 3. Notice that within the integral we have tweets to the power of TDT so we can rearrange this expression and show that DU divided by eon of 3 or 1 divided by eon of 3DU. So 1 divided by ln of 3 d u is equal to 3 to the power of t dt. In other words, we're multiplying both sides by dt and we're dividing both sides by ln of 3. This is how we arrive at this expression. So now the integral can be rewritten as sequence of you. And 3 it's the power of TDT becomes. 1 divided by Eon of 3 DU, we can just write DU. And we can factor out the constant 1 divided by ln of 3. And this is now an elementary integral, so we get our constant 1 divided by ln of 3, and the integral of 2 is ln of the absolute value of. 2 U plus The ancient EU. And we're going to add a constant of integration because that's. An indefinite integral. The final step is to replace you with. 3 to the power of t, so we get a 1. Divided by Ellen of 3, or we can just write a combined fraction, right, we can just write Ean of 3. In the denominator. And in the numerator we have ln of the absolute value of 2. Of 3 to the power of t + Tangent of 3. So the power of t. And we're going to add a constant of integration C. That's our final answer and thank you for watching.

Evaluate the integrals in Exercises 37–44.∫ eᵗ √[tan²(eᵗ) + 1] dt

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Evaluate the integrals in Exercises 37–44.
∫ eᵗ √[tan²(eᵗ) + 1] dt