Evaluate the integrals in Exercises 23–32.
∫₀^(π/6) √(1 ...

Question

Evaluate the integrals in Exercises 23–32.

∫₀^(π/6) √(1 + sin(x)) dx

(Hint: Multiply by √((1 - sin(x)) / (1 - sin(x))))

Accepted Answer

Hello there, today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Compute the integral evaluated from 0 to pi divided by 6 of the square root of 1 minus sin of x dx. OK. So it appears for this particular prom, we're asked to compute, or rather solve for the integral that is provided to us by the prom itself. So now that we know that we're ultimately trying to solve for this integral, our first step that we need to take is we need to multiply the integrand by the square root of 1 + sin of x divided by 1 minus sin of x, and when we multiply the integrand by this value. We will find that we can write that the square root of 1 minus sin of x multiplied by the square root of 1 + sin of x divided by 1 minus sin of x. Will be equal to the square root of 1 minus sin square of x divided by the square root of 1 + sin of x. Fantastic. So now we need to simplify the numerator by recalling and using the Pythagorean identity. And when we do, we will find that we're able to write the square root of 1 minus sin to the power 2 x divided by the square root of 1 plus sin of x will be equal to. The square root of cosine square of x divided by the square root of 1 + sin of x, which will be equal to the absolute value of cosine of x. Divided by the square root of 1 + sin of x. And we need to note and recall that since cosine of x is greater than or equal to 0. On the closed interval, which we use square brackets via square brackets denote a closed interval, and we need to note that since cosine of x is greater than or equal to 0 on this closed interval of 0, pi divided by 6. That will mean that undoubtedly, that the absolute value of cosine of X will be equal to cosine of x. So the absolute value of cosine of X will be equal to cosine of x. So based on this information, we can therefore write that the integral evaluated from 0 to pi divided by 6. Will be So once again, the integral evaluate from 0 to pi divided by 6 of the square root of 1 minus sin of xDX will be equal to the integral evaluated from 0 to pi divided by 6 of. Cosine of x divided by the square root of 1 plus sin of x. Fantastic, and don't forget. To multiply by DF X. Awesome. So Drum roll here we now need to recall and use the substitution method. And when we recall and use the substitution method for this particular problem, we could write that u is equal to 1. Plus sin of x and this will mean therefore that then this will mean. That DU, so we could say that then we can safely write that. DU by DX will be equal to cosine of X, which will mean that DU will be equal to cosine of XDX. Fantastic. Moving right along here. So, We can now at this stage, rewrite our limits, and when we do, we'll be able to write that X is equal to 0 will mean that U is equal to 1 plus sin of 0 because we're plugging in a value of 0 in for x, and this will be equal to 1 + 0, which is equal to 1. And when X is equal to pi divided by 6, that will mean that you will be equal to 1 plus sin of pi divided by 6, which will be equal to 1 + 1/2, which will be equal to 3 divided by 2. Fantastic. And now we need to substitute these new variables into our integral. So we need to take the integral evaluated from 0 to pi divided by 6 of cosine of x divided by the square root of 1 plus sin of x dx. Will be equal to the integral of 1. 2, so once again the integral evaluated from 123 divided by 2, our new limit values of DU. Divided by the square root of you. Which will be equal to the integral. Which is evaluated from 123. Divided by 2. Of u to the power of -1/2 d. And now At this stage, we need to integrate by recalling and using the power rule which states that the integral of u to the power of n. D is equal to u to the power of n + 1 divided by n + 1 plus c where c denotes some constant variable. And we need to know where the where this variable n. Cannot equal -1. So once again, the intervaln cannot, or actually I should say specifically, the variable N cannot equal -1. So with that said. We can go ahead and write that the integral evaluated from 1 to 3 divided by 2 of u to the power of -1/2 duu will be equal to u to the power of 1/2 divided by 1/22, evaluated from 1 to 3 divided by 2. Which will be equal to 2 multiplied by the square root of u. Which is evaluated from 123. Divided by 2. Which will mean that this is equal to 2 multiplied by the square root of 3 divided by 2 minus the square root of 1 in parentheses, which when we evaluate this expression by plugging it into our calculator and we will find that it's equal to the square root of 6 minus 2, and that is our correct and final answer hooray, we did it. So going back up to the top to look at the prom itself to recap what we've learned, we can safely say without a doubt that our computed integral, final answer without a doubt, once again will be the square root of 6 minus 2. And that's it. That is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Evaluate the integrals in Exercises 23–32.∫₀^(π/6) √(1 + sin(x)) dx(Hint: Multiply by √((1 - sin(x)) / (1 - sin(x))))

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Evaluate the integrals in Exercises 23–32.
∫₀^(π/6) √(1 + sin(x)) dx
(Hint: Multiply by √((1 - sin(x)) / (1 - sin(x))))