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

Question

9–61. Trigonometric integrals Evaluate the following integrals.

38. ∫ tan⁵θ sec⁴θ dθ

Accepted Answer

Hello. In this video we are going to be evaluating the following integral. So we are given the integral of tangent cubed of X multiplied by skin to the 6th power of XD X. Now the first thing we want to do is we want to go ahead and simplify the integrand into terms of tangent. So what we can do is we can first reduce the power of seek into the 6th. If we raise 2 to the third power, then we can rewrite the integral to be the integral of tangent cued of X multiplied by sequent squared of X raised to the 3 power d X, and if we were to use the Pythagorean identity for seek square of X, we can rewrite this integran to be the integral of tangent cubed of X multiplied by the quantity. Of 1 plus tangent square of X raised to the third power DX. Now, what we can go ahead and do here is we can simplify the integram further by using a U substitution. If we allow you to equal the tangent of X, then D U will equal to squared XD X, but we don't have a squad in our integrand. So instead, let's go ahead and rewrite C squared into terms of tangent for our DU statement. First, we will go ahead and divide both sides of this statement by seek squared since we are solving for DX, and that will leave us with DU. Divided by 2 X is equal to DX. By applying the Pythagorean identity for Ckin squared, we can rewrite the denominator as DU divided by 1 plus tangent squared of X, which is equal to DX. Now, since we stated that U is equal to tangent of X, we can further rewrite this as DU divided by 1 plus U squared, and this will equal to DX. So, if we apply the substitution of DX and substitute every tangent in the integran with our U substitution, then we can further simplify this integral to be the integral of u cubed, multiplied by the quantity 1 plus U squared raised to the third power, multiplied by DU divided by 1 + u squared. Now notice that we are multiplying a 1 + square cubed by 1 divided by 1 + u squad. We can further simplify this by eliminating one power of the quantity 1 + u squared, and that will allow us to rewrite the integral to be the integral of u cubed, multiplied by 1 + U squared, quantity squared DU. Now what we are going to do is we are going to simplify using algebraic methods. Let's go ahead and expand one plus U squad, raise the squared power. That will leave us with the integral of U cubed multiplied by the quantity, 1 + 2 U squared, plus the 4th power DU. And if we were to distribute U cubed into every term in the integrand, that will allow us to further rewrite the integral as you cubed. Plus 2 multiply by u to the 5th power plusu raise to the 7th power d u. All we need to do now is we need to take the antiderivative of each of the terms in the integrand. By taking the antiderivative, the integral will give us 1/4 u to the 4th power plus 1/3 u to the 6th power. Plus 1/8 to 8th power, and since we are working with a general integral, we will add a general constant C. Now, all we need to do is we need to resubstitute you back into terms of X. Now, earlier we stated that U is equal to tangent of X, and so we are just going to resubstitute you with tangent instead. That will allow us to rewrite the integral to be 1/4 multiplied by tangent to the 4th power of X plus 1/3 multiplied by tangent to the 6th power of X plus 1/8 multiplied by tangent to the 8th power of X plus the general constant C, and this is going to be the solution to the given integral. And with that being said, the overall solution is going to be D. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

9–61. Trigonometric integrals Evaluate the following integrals.
38. ∫ tan⁵θ sec⁴θ dθ

Key Concepts

Trigonometric Functions

Integration Techniques

Secant and Tangent Identities

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