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

Question

9–61. Trigonometric integrals Evaluate the following integrals.

20. ∫ sin⁻³ᐟ²x cos³x dx

Accepted Answer

Hello. In this video, we are going to be evaluating the given integral. We are given the integral of sin to the half power X multiplied by cosine to the 5th power of XDX. Now, we want to go ahead and rewrite any trigonometric functions with odd-powered exponents into trigonometric functions that have even powered exponents. That way, we may be able to apply Pythagorean identities. So when we are looking at this given integral, the first thing we want to reduce is the cosine to the 5th power. In order to do this, we can rewrite cosine to the 5th power as a product of cosine functions. So, we can rewrite the given integral to be the integral of sin to the negative half power of X multiplied by cosine to the 4th power of X multiplied by cosine of XDX. Now, cosine to the 4th power can be rewritten as cosine squared to the power of 2. So we can further rewrite the integral to be signed to the negative half power of X, multiplied by cosine squared of X raised to the second power multiplied by cosine of XD X. Now, if we were to apply the Pythagorean identity for cosine squared of X, we can further rewrite the integral to be signed to the negative half power of X multiplied by 1 minus sine squared. Of X raises to the second power multiplied by cosine of XDX. Now, let's go ahead and simplify this function or this integral by using a U substitution. Now, the problematic term here is going to be the sign to the negative half power. So let's go ahead and allow you to equal to sin of X. That means that if we want to find a substitution for DX, we will take the derivative of this U substitution, and we will get DU is equal to cosine of XDX. We have a cosine of XDX term in our integral, and that's going to be substituted with DU, and every sine function within the integral will be substituted with U. So by using this U substitution, we can simplify the integral to be the integral of U to the negative half power multiplied by 1 minus U2, raise the second power, DU. Now what we need to do is we need to simplify this by using algebraic methods. We will start by expanding 1 minus U2 to the second power, and that will leave us with U to the negative half power multiplied by 1 minus 2u2 plusu to the 4th DU. We will then distribute you to the negative half power into that quantity, and that will leave us with you to the negative half power. Minus 2 multiplied by U to the three halves, plus U 7 halves, and by taking the antiderivative of each term, we will be left with 2 multiplied by U half power, plus 2 multiplied by the antiderivative of u to the 3 halves, which is going to be 2/5 U 5 halves. Plus the antiderivative of the last term, which is going to be 2 divided by 9, multiplied by U to the nine halves power. And since this is an indefinite integral, we'll add a generic constant C. Now, all we need to do is we need to backs substitute you into terms of X. So we will be resubstituting sin of X for every U term and multiplying these coefficients together, and that is going to give us a final solution of 2 multiplied by sin to the power of X. Plus 4/5 multiplied by sine to the 5 halves power of X plus 2 divided by 9 multiplied by sine to the 9 halves power of X plus C. and this is going to be the simplest form of the overall solution to the integral, and with that being said, the overall solution is going to be C. 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.
20. ∫ sin⁻³ᐟ²x cos³x dx

Key Concepts

Trigonometric Identities

Integration Techniques

Substitution Method

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