3. Describe the method used to integrate sin³x.

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Hello. In this video, we are going to compute the integral of sin to the 5th power by using the reduction method and substitution. So, since we are given an odd-powered exponent for sign, what we want to do is we want to reduce this version of the integral into a form where we have even powered values of sign. That way we can be able to apply Pythagorean identities. So, in order to approach this problem, the first thing we can do is we can rewrites sign to the 5th power as sign to the 4th power of X, multiplied by sine of XD X. Now, sine to the 4th power of X is equivalent to sine squared raised to the power of 2. So, if we were to rewrite this again, we can rewrite sine to the 4th power to be signed to the power of 2 quantity squared multiplied by sine of XD X. Furthermore, we can apply the Pythagorean identity for sin squared of X in this case. Now, the Pythagorean identity for sine squared is going to be 1 minus cosine squared of X. So, that will allow us to rewrite the integral to be 1 minus cosine squared of X, this quantity raised to the second power minus multiplied by sin of XD X. Now, in this case here, if we were to allow a substitution U to equal to cosine of X, Then DU is equal to negative sign of XDX. And by moving the negative sign to the left hand side, we will end up with negative DU is equal to sin of xDX. So we are able to substitute this extra factor of sine to be negative DU, and our cosine function will be substituted with U. This will allow us to simplify the function to be the integral of 1 minus u squared, that quantity squared multiplied by negative d u, which we will go ahead and move the negative to the front of the integral since it is a constant value. Now we will go ahead and simplify the integral by using algebraic methods. So if we were to expand the quantity 1 minus u2 to the power of 2, we can rewrite the integran to be negative integral of 1 minus 2u2 plusu the fourth power DU. And by taking the antiderivative of each term, we will end up with -1 multiplied by U. Minus 2/3. You to the 3 power. Plus 15th to the 5th power and since we are working with the indetermined integral we will add on a general constant c. Now what we will go ahead and do is we will go ahead and distribute that negative sign. That will leave us with negative U plus 2/3 U3 minus 1/5 U 5 plus C. And finally, all we need to do is just resubstitute you back in terms of X, which our original substitution was is equal to cosine of X. That will leave us with negative cosine of X plus 2/3 cosine to the third power of X minus 1/5 multiplied by cosine to the 5th power of X plus C. And this is going to be the simplest form of the value of the given indetermined integral. And with that being said, the solution to this problem is going to be a. 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.

3. Describe the method used to integrate sin³x.

Key Concepts

Integration Techniques

Trigonometric Identities

Substitution Method

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