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

Question

9–61. Trigonometric integrals Evaluate the following integrals.

19. ∫[0 to π/3] sin⁵x cos⁻²x dx

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says to evaluate the integral, and we're given the integral from 0 to pi divided by 4 of sine cub to X, multiplied by cosine to the minus 2 of X D X. So our first step to evaluate this integral is to rewrite or integrand as a fraction. So we'll have the integral from 0 to pi divided by 4 of sinecued of X divided by cosine squared of X, multiplied by DX. Now, the next step is to rewrite our numerator as sin of x multiplied by sin squared of X. So this is going to be equal to the integral from 0 to pi divided by 4 of sin of X. Multiplied by and instead of sin square x, we'll use our trig identities to rewrite that as the quantity of 1 minus cosine squared of X in quantity, and that's going to be divided by. Cosine squared of X, that's all multiplied by DX. Now we just need to distribute our sin of Xs in the numerator, but this is going to be equal to the integral from 0 to pi divided by 4 of sin of X. Minus cosine squared of X multiplied by sin of X. And that whole quantity is going to be divided by. Cosine squared of X, and that's all multiplied by DX. So now we'll break our single fraction up here into two separate fractions, which will give us two separate integrals. So this is going to be equal to the integral from 0 to pi divided by 4 of sin of X. Divided by Cosine squared of X. DX Minus the interval from 0, you pi divided by 4 of, and for the second term, we'll have cosine square to Xx divided by cosine square of Xx, which is 1, and then that's multiplied by. Sign of X DX So now we just need to integrate each one of these integrals separately. Now, for the first integral, we'll need to use substitution to evaluate it. So we're going to set U equal to cosine of X, and that means that DU is going to be equal to minus sine of X. DX But if I look at my integral, I just have sin of XDX, so we need to multiply both sides of this equation by -1, and so we'll have minus DU is equal to sin of X. DX. So, that means our first integral is going to be equal to the integral, and here we have to change our limits of integrations, since we've changed variables. And so, when X is equal to 0, U is going to be equal to cosine of zero, which is 1, so we'll have 1 for a lower limit. And when X is equal to pi divided by 4, U is going to be equal to cosine of pi divided by 4, which is going to be 1 divided by the square root of 2. We'll have that value as the upper limit. Now, for our integran, we will have in place of sin of XDX minus DU, and that's going to be divided by in the place of cosine squared X, we'll have U2d. Then we'll have minus and for our second integral here, we can actually integrate that directly since we can integrate sine of X, which will give me minus cosine of X, and this needs to be evaluated from 0 to pi divided by 4. So, continue to evaluate this expression. This is going to be equal to, and we can actually integrate our first integral here. So we'll have minus, and then when we integrate 1 divided by U squared with respect to you, we get minus 1 divided by U. Then this needs to be evaluated from 1 to 1 divided by the square root of 2. Then we'll have plus, and here we need to substitute in our upper and lower limits. Into cosine of X, so we'll have cosine of pi divided by 4 minus cosine of 0. And continuing to simplify this expression, this is going to be equal to. When we plug in our upper and lower bounds into our expression, we just have 1 divided by you, and when we substitute in U equal to 1 divided by the square 2, or upper bound, we just get the square root of 2. Then we'll have minus, we substitute in 1 for you, we just get 1 for our lower bound. Then we'll have plus, the cosine of pi divided by 4 is the square root of 2 divided by 2. Don't have minus cosine of 0, which is 1. And now just collecting like terms and simplify this expression, this is going to be equal to 3 multiplied by the square of 2, divided by 2 minus 2. And this here is the answer that we were looking for for this problem. So, I hope that this explanation has been useful, and I'll see you in the next video.

9–61. Trigonometric integrals Evaluate the following integrals.
19. ∫[0 to π/3] sin⁵x cos⁻²x dx

Key Concepts

Trigonometric Functions

Integration Techniques

Definite Integrals

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