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

Question

9–61. Trigonometric integrals Evaluate the following integrals.

43. ∫ tan³(4x) dx

Accepted Answer

Welcome back, everyone. It's another video. Evaluate the integral of tangent cubed to XDX. For this problem, first of all, what we're going to do is rewrite our integral as the integral of tangent of 2 Xs multiplied by tangent squared of 2 X. In other words, we are rewriting the cube. As the product of two tensions, one raises the power of one and another squared, right? And now let's analyze the engine squared. Well, essentially we can rewrite the engine squared of 2 X using a trigonometric identity, and that'll be squared of 2 X minus 1. So our integral becomes tensions. Of 2 X multiplied by squad of 2 X minus 1 D X. From here, we're going to introduce substitution. Let's suppose that you equals tangents. Of 2 X, then the derivative DU divided by D X is equal to c 22 X, and according to the chain rule, we're multiplying by the derivative of 2 X, which is 2. So basically 2c can squared of 2 X. Now, going back to our integral, we can actually distribute and show that it is equal to the difference between two integrals. So, tangent. Of 2 X multiplied by squad of 2 X D X minus the integral of tangent of 2 X. And now let's go ahead and solve for each integral. So now let's consider the first integral. As we can see, it contains squad of 2 X multiplied by DX. So if we rearrange our derivative, we can show that 12 DU. Is equal to c2 of 2 XDX. So for the first integral, if we analyze it separately, we have tangent of 2 X which is U, multiplied by squad of 2 XDX, that's 1/2 DU, so we can factor out 12, and we can now see that we're integrating UDU. According to the power rule, that will be used squared divided by 2, and we add a constant of integration C. Combining the constants, we end up with 14th. You squared. Plus C. Now we can substitute for you and that'll be 1/4 tangent squared of 2 X plus C. And now we have our 2nd integral. It will also have a constant of integration, so let's label the constant of integration for. The first integral is C1, so that we could later on differentiate between those constants. Now, considering the integral of tangent of 2 X, we can just express tangent of 2X as Sign of 2 x divided by cosine of 2 X. And now what we can do from here is basically introduce a substitution. Let's suppose that you equals. Cosine of 2 X. And then we can show that the you. Divided by D X is equal to the derivative of cosine of 2 X, and that's negatives of 2 X multiplied by the derivative of 2 X, which is basically 2, right? So now we can show that sign. Of 2 X multiplied by the X is equal to -12 DU. And then our integral becomes. 12, we can factor out the constant DU because we have sin of 2 XDX in the numerator, so we are going to introduce integral of DU divided by U. And now this is a basic integral that would be -12 LN of the absolute value of you. Plus C2. And this is 1/2 of the absolute value. Of cosine of 2X because this is our. Second substitution, right? We have our U equals cosine of 2 X, and then we add our constant of integration C2. And now we can evaluate our main integral, right? Going back to that main integral, we can just label it as. I And we can now show that i equals our first integral. Let's recall that it was equal to. 14th tangent squared of 2 x plus C1. For now, let's forget about C1. Let's remember that we're subtracting the second integral which has a negative value. So this would result in a positive sign, and this means we're adding 1/2. LN of the absolute value of cosine of 2 X, and then we can combine our constants C1 and C2. So we're basically adding C in a general case. And that would be our final answer for this problem. Thank you for watching.

9–61. Trigonometric integrals Evaluate the following integrals.
43. ∫ tan³(4x) dx

Key Concepts

Trigonometric Functions

Integration Techniques

Definite and Indefinite Integrals

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