Evaluate the integrals in Exercises 53–76.
61. ∫(from 0 ...

Question

Evaluate the integrals in Exercises 53–76.

61. ∫(from 0 to 2)dt/√(8+2t²)

Accepted Answer

Welcome back everyone. Evaluate the definite integral from 0 to 1 of dx divided by square root of 5 + 3 x 2. For this problem, we're going to use one of the integrals, and specifically that's the integral of Dx divided by the square root of x2 + A2. Now this integral is equal to ln of the absolute value of x + square root of x2 + a2 plus a constant of integration c. So in this case, the main difference is that instead of x2 we have 3 x 2. So what we can do is simply rewrite the original form of the integral, that's the integral from 0 to 1 of the x. Divided by square root of 3 x 2 plus 5, we can basically factor out 3 in the denominator so we're going to get the integral from 0 to 1 of d x divided by. square root of 3 x2 + 5/3. Now we can basically separate the radical terms to get 1 divided by a square root of 3. We can also factor out the constant, right, and get the integral from 0 to 1. Of d x divided by square root of x2 plus 5/3, notice that by analogy we can then show that a squared is equal to 5/3. So A is going to be the square root of 5/3. Remember that we're taking the positive solution here and we can also write it as. The square root of 5 divided by the square root of 3. Which is equal to the square root of 15 divided by 3, we can rationalize the denominator if we multiply both sides by the square root of 3. Well done, so we have a 2 and A. We can just apply the formula. Let's call it integral i, and integral i is going to be equal to ln of the absolute value of. So what we're doing is taking X. Let's also add a constant of. The constant in front of the integral, right? So that's 1 divided by the square root of 3, and now we can apply the formula, right? So we're going to get lm of the absolute value of x plus square root of. X2 + A2, so that's X2 + 5/3. Notice that x is between 0 and 1, so there is no necessity to use the absolute value. We can just include parenthes, and we want to evaluate the result between 0 and 1. So let's go ahead and apply the fundamental theorem of calculus. Part two, we can also rationalize the denominator once again, so that'd be square root of 3 divided by 3. We can factor out that constant. And now let's go ahead and evaluate the natural log at each limit. We're going to get LN of 1 plus. Square root of 1 squad, so that's 1 + 5/3. And we want to subtract. LM off. 0 + square root of 0 + 5/3, so that's minus ln of. Square root of 5/3. Well then, now what we can do is simplify. Let's go ahead and simplify, so that would be the square root of 3 divided by 3. In parenthesis we have Ellen of. 1 + square root of. Now 1 + 5/3 is going to be square root of. 8/3. Minus aen of Square root of 5/3. We can now use the properties of logarithms to simplify further. This is going to be equal to the square root of 3 divided by 3. Alan Of. In the numerator we have one plus. square root of 8 divided by 3, we can write it as square root of 8 divided by the square root of 3. Now, square root of 8 is equal to 2 square root of 2, and now we can also rationalize the denominator, right? So, multiplying both sides by the square root of 3, we get 2 square root of 6. Divided by 3. And then since we're subtracting ln off the square root of 5/3, we are basically dividing by square root of 5. Thirds, which can be written as square root of 5 divided by square root of 3. Again, we can rationalize the denominator that's going to be square root of 15 divided by 3. And now let's find the common denominator. In the numerator, we will have 1, which is 3 divided by 3, right? So that's 3 plus 2 square root of 6 divided by 3, where 3 is the common denominator. And the reason why this is useful is simply because now we can cancel out each denominator. It is 3 and 3, right, and simplify the answer. So we are going to get the square root of 3 divided by 3 ln of. 3 + 2 square root of 6 Divided by a square root of 15. Well done, and now we can once again. Rationalize the resultant. Denominator, we can multiply both sides by square root of. 15. Which will give us square root of 3 divided by 3 ln of. 3 multiplied by a square root of 15 is going to be 3 square root of 15 plus. 2 multiplied by the square root of 15 multiplied by square root of 6 is going to give us 2 square root of 90. And we are going to divide it by 15. Additionally, we can rewrite the square root of 90s, square root of 9 multiplied by square root of 10. And square root of 9 gives us 3, so 2 multiplied by. 3 is going to give us 6. And now from here. We can additionally divide both sides by 3, right? So let's go ahead and do that to get our final answer, square root of 3 divided by 3. Alan of The square root of 15 + 6 divided by 3 gives us 2. Square root of 10. Divided by 15 divided by 3 is 5. This is how we arrive at the final answer for this problem. Thank you for watching.

Evaluate the integrals in Exercises 53–76.61. ∫(from 0 to 2)dt/√(8+2t²)

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Evaluate the integrals in Exercises 53–76.
61. ∫(from 0 to 2)dt/√(8+2t²)