Evaluating integrals Evaluate the following i...

Question

Evaluating integrals Evaluate the following integrals.

∫₋π/₂^π/² (cos 2𝓍 + cos 𝓍 sin 𝓍 ― 3 sin 𝓍⁵) d𝓍

Accepted Answer

Welcome back, everyone. Compute the integral from negative pi divided by 4 up to pi divided by 4 of cosine 6 X plus sine x cosine x minus 4 sine cubed of XD X. A says 1/3, B 1/3, C 16, and D 16. So for this problem, let's begin by rewriting our integral as 3 integrals, right? Because we have a sum and a difference. So we can essentially say that our first integral has cosine of 6 X. Let's label this as F of X for our integrant. For the second part, we have sin x, cosine X. Let's label this function as G of X. And for the third integral, we have 4 sine cubed of X. So let's label it as H of X. Because our limits of integration involve opposite numbers, we want to use symmetry to evaluate each integral. So let's consider the symmetry of each function, starting with F of X, we want to evaluate F of negative X, which is cosine of -6 X. And that we know that cosine is an even function, which means that cosine of negative X is equal to cosine of X. So, cosine of -6 X is equal to cosine of 6 X. And therefore F of negative X is essentially equal to F of X. There are 4 F of X is and even function. Similarly, we can evaluate G of negative X, which is. Sign of negative X multiplied by cosine of negative X. Now sign is an odd function. Which means that sign of negative X is equal to negative sign of x. And for a cosine, we already know that cosine of negative X is, cosine of X. So this is equal to negative G of X because we have a negative sign in front. And therefore G of X is an odd function. And finally, we want to consider each of negative X, which is 4 sine cubed of negative X. This is equal to 4 multiplied by sin of negative X. Cubed We can rewrite it as 4 multiplied by. Negative sign. Cubbed of X because we already know that sign of X is an odd function, so we can basically take out the negative sign, right? And when we're cubing a negative number, we basically get that negative sign in front because we can essentially cube a positive number and add a negative sign at the end, right? So this is equal to -4 cubed of X. Now, because we get -4 sine cubed of X, we have shown that H of negative X is equal to negative H of X. Which also means that H of X is an odd function. And now let's recall the properties of integrals. If we have odd functions, the integral from a negative constant up to the same positive constant, meaning an opposite number, is equal to 0. Therefore, the 2nd integral is equal to 0, since G of X is an odd function, and the 3rd integral is also 0 because H of X is an odd function. So the only integral that we want to take care of is the integral from negative pi divided by 4 up to pi divided by 4 of cosine. 6 XD X. Now, the integral of cosine is sign, so we first will get sine of 6 X. And because we have a linear function for cosine, it's cosine of 6 X, according to the reverse chain rule, we have to divide by the slope of the linear function. So we are dividing by 6, right? Meaning our integral is sin of 6 X divided by 6. And we want to evaluate it from negative pi divided by 4 up to pi divided by 4. And now let's go ahead and apply the fundamental theorem of calculus part 2. We can also factor out 1/6, and then imprecies we're evaluating sign. Of 6 pi divided by 4, which is of 3 pi divided by 2 minus. Syme Of -3 pi divided by 2. And now let's go ahead and simplify the result. Specifically, we want to recall that sin of 3 pi divided by 2 is equal to -1, so we got 16 in -1 minus. Now, since sine is an odd function, we can rewrite sine of -3 pi divided by 2s, negative sign of 3 pi divided by 2. So we are basically adding sign of 3 pi divided by 2, or basically adding -1. 1/6 multiplied by -1 + -1 is equal to -2 divided by 6 or 13. So the correct answer for this problem corresponds to the answer choice A-1/3. Thank you for watching.

Evaluating integrals Evaluate the following integrals.

∫₋π/₂^π/² (cos 2𝓍 + cos 𝓍 sin 𝓍 ― 3 sin 𝓍⁵) d𝓍

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Evaluating integrals Evaluate the following integrals.∫₋π/₂^π/² (cos 2𝓍 + cos 𝓍 sin 𝓍 ― 3 sin 𝓍⁵) d𝓍

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Evaluating integrals Evaluate the following integrals.

∫₋π/₂^π/² (cos 2𝓍 + cos 𝓍 sin 𝓍 ― 3 sin 𝓍⁵) d𝓍