7–84. Evaluate the following integrals.
35. ∫ from 0 to ...

Question

7–84. Evaluate the following integrals.

35. ∫ from 0 to π/4 [(tan²θ + tanθ + 1) sec²θ] dθ

Accepted Answer

Welcome back, everyone. Evaluate the integral. Integral from 0 to pi divided by 3 of tent squared the minus 2 tangent theta plus 5, all multiplied by c2 theta theta. For this problem, we're going to begin with U substitution. Let's suppose that U equals tangent theta, then the derivative DU divided by the theta is going to be equal to the derivative of tangent, which is squad theta. Now, we are going to identify the new limits of integrations, starting with the lower limit. Our U1 is going to be tangent of 0, which is 0. In the upper limit of integration, U2 is going to be tangent of pi divided by 3, which is square root of 3. And then we can rewrite our integral in terms of you. We begin with our limits, that'd be 02 square root of 3, and we want to integrate U2 minus 2 U + 5, and 2 squared the the theta is basically the U, right, because we can multiply both sides of our derivative by the theta. And then we can split it into 3 integrals, right? We get integral from 0 to square root of 3 of U squared DU. Minus 2 integral of UDU and our limits of integration would be the same from 0 to square root of 3, plus 5 integral from 0 to square root of 3DU. Now let's apply the power rule for the first one we get you cubed divided by 3. For the second one we get you squared divided by 2. And 2 divided by 2 gives us 1, we can just say that we get minus U squared, and finally plus 5u because the integral of the U is U. And we want to evaluate this result from 0 to square root of 3. So let's substitute square root of 3 for you, which gives us square root of 3. Cubed divided by 3. Minus square root of 32 + 5 square roots of 3, and we're subtracting the value of the result at 0, which is basically 0 because every term contains U, right? So now let's simplify square root of 3 cubed is going to be equal to 3 square roots of 3. And we're dividing that by 3, so we can just cancel out the 3. We're subtracting 3 and adding 5 square roots of 3. Which is equal to square roots of 3 + 5 square roots of 3, that's 6 square roots of 3 minus 3, and that'd be our final answer for this problem. Thank you for watching.

7–84. Evaluate the following integrals.
35. ∫ from 0 to π/4 [(tan²θ + tanθ + 1) sec²θ] dθ

Key Concepts

Integration

Trigonometric Identities

Substitution Method

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