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

Question

9–61. Trigonometric integrals Evaluate the following integrals.

13. ∫ sin⁵x dx

Accepted Answer

Hello. In this video, we are going to be evaluating the following indetermined integral. So we are given the integral of sin to the 7th power of XDX. Now, whenever we approach a trigonometric integral with an odd powered exponent, we want to rewrite this into an even powered exponent so that we can apply Pythagorean identities. The first thing we can do is we can rewrites signed to the 7th power as signed to the 6th power. We can write the integral as signed to the 6th power of X, multiplied by sine of XD X. Now, sine to the 6th power is equivalent to sine squared race to the 3rd power. So we can rewrite that term to be sine squared of X raised to the third power, multiplied by sin of XD X. By using the Pythagorean identity, we can replace sine squared of X as 1 minus cosine squared of X, and so we will have the integral of 1 minus cosine squared of X raises to the third power multiplied by sin of XD X. Now let's go ahead and apply a U substitution. If we allow you to equal to cosine of X, then D U is equal to negative of X D X, and by applying the substitution, we can rewrite the integral to be the negative integral of 1 minus u squared cubed, and since we have sin of XDX in the integral already, that will be substituted with DU. So, now what we need to do is we just need to simplify using algebraic methods. We will go ahead and expand 1 minus U2 cubed. That is going to give us the negative integral of 1 minus 3 U 2. Plus 3 to the 4th power. Minusu to the 6th power d u and by taking the antiderivative of each term, we will have the negative u m plus u cubed. Oh, minus You cubed, apologies. Plus 3/5, U 5 minus 1/7, Uth power, and since we were working with an indefinite integral, we will add a generic constant C. Now what we're going to do is distribute the negative sign. We will have negative U plus U cubed minus 35 U 5th power, plus 1/7 U 7th power plus C. And now what we need to do is we just need to substitute you back in the terms of X. That is going to leave us with negative cosine of X, plus cosine cubed of X. Minus 3/5 multiplied by cosine to the 5th power of X, plus 1/7 multiplied by cosine to the 7th power of X plus C. and this is going to be the final form of the solution to this problem. And the overall solution is going to be B. 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.
13. ∫ sin⁵x dx

Key Concepts

Trigonometric Functions

Integration Techniques

Power Reduction Formulas

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