9–61. Trigonometric integrals Evaluate the following integrals...

Question

9–61. Trigonometric integrals Evaluate the following integrals.

53. ∫ from 0 to π/4 of sec⁴θ dθ

Accepted Answer

Welcome back, everyone. Evaluate the integral from 0 to pi divided by 6 of sequence to the power of 4 of XD X. For this problem we're going to begin with one of the identities in trigonometry, and that the second squad of X is equal to 1 plus tangent squared of X. Now what we have to do is simply introduce a substitution and let's suppose that u equals tangent of X. Let's recall that the derivative of tangents, the divided by DX is equal to squad of X, right? So squad of X multiplied by the X is going to be our DU. And then we can rewrite our integral. For now, let's not worry about the limits of integration. Let's just focus on sequent to the power of 4 of XDX, right? So that would be sequent squad of X. Multiplied by sequent squared of XDX, right? We initially had limits from 0 to pi divided by 6, and now we're going to convert that in terms of you. So sequence squared XDX is going to be our DU. And squared X can be rewritten as 1 + tent squared X, which is U2, right? Now, our goal is to change the limits of integration in terms of you. So the lower limit of integration U1 is going to be equal to tangent of 0, which is 0. And the upper limit of integration is going to be tangent of pi divided by 6. And this is equal to square root of 3. Divided by 3. Meaning our integral becomes integral from 0 to square root of 3 divided by 3 of 1 plus U 2 DU. And then we can integrate the integral of the U is just you. The integral of you squared is u cubed divided by 3. And we want to evaluate this definite integral from 0 to square root of 3 divided by 3. This is going to be equal to square root of 3 divided by 3. Plus 1/3, we can seek out 1/3 multiplied by square root of 3 divided by 3 cubed. At 0, both terms are 0, so we can just evaluate the result immediately. We can factor out square root of 3 divided by 3. In paries, we get 1 plus 1/3. Multiplied by Square root of 3 divided by 3 squared, right? And now let's simplify. So, we get square root of 3 divided by 3 in 1 plus 1/3 multiplied by 3 divided by 9. That's 1/3, right? Which is equal to square root of 3 divided by 3 in 1 plus 19th. Which is square root of 3 divided by 3 multiplied by 9 divided by 9 plus 1 divided by 9, that's 10 divided by 9. And finally, let's multiply. This is going to be equal to 10 square root of 3 divided by 27, which is our final answer for this problem. Thank you for watching.

9–61. Trigonometric integrals Evaluate the following integrals.
53. ∫ from 0 to π/4 of sec⁴θ dθ

Key Concepts

Trigonometric Functions

Integration Techniques

Definite Integrals

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