7-56. Trigonometric substitutions Evaluate the following integ...

Question

7-56. Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.

54. ∫ y⁴/(1 + y²) dy

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to evaluate the integral using trigonometric substitution, and we're given the integral of 2 Z 4th divided by the quantity of 1 plus Z squared in quantity DZ. So we're asked to evaluate this interval using trigonometric substitution. So we were given the interval of 2 Z 4th, divided by the quantity of 1 plus Z squared and quantity D Z. So to use trion message substitution, we want to set Z equal to tangent of the. And that means that Dz is going to be equal to. Sequence squad of theta theta. So making this substitution, we see that our interval is going to be equal to the interval of 2 multiplied by tangent to the fourth of theta multiplied by squared of theta, and that's going to be divided by the quantity of 1 plus tangent squared of theta in quantity the theta. So now we just need to simplify this expression. And so this is going to be equal 2. We have a constant of 2 in the numerator that we can factor out and pull in front of the integral. So we'll have 2 multiplied by the integral and in the numerator we'll have tangent to the 4th of the multiplied by sequence squared of theta, and that is going to be divided by, and here we call that 1 plus tangent square of theta is equal to sequence of theta. So in the denominator we'll have sequence squared of theta, and then that fraction gets multiplied by the theta. Now the 2 squared of theta terms will cancel out. And what we're going to be left with is 2 multiplied by the integral of tangent to the 4th of theta the theta. And so we want to do is rewrite tangent to the 4th of theta as the quantity of tangent square theta in quantity squared. And so doing so this is going to be equal to 2 multiplied by the integral of the quantity, and in place of 10 squad theta, we will rewrite it as sequence square theta minus 1 using our trig identities in quantity, and then that quantity is going to be squared, and that's multiplied by the theta. So now we want to do is multiply everything out in the integrand. In doing so, this is going to be equal to 2 multiplied by the integral of the quantity of sequent to the 4th of theta minus 2 sequent squared. Plus 1 in quantity d theta. So now what we want to do is to separate this out into 3 separate integrals. In doing so, this is going to be equal to. 2 multiplied by the integral of 2 to the 4th of theta d theta. Minus and here we have a factor of 2 multiplied by 2 as well. So we'll have 4. Multiplied by the integral of sequence squared of data. The theater Plus 2 multiplied by the integral of the data. Now, the last two integrals are simple enough to do directly. However, we will need to actually do a little bit of work to evaluate that first integral. So we're going to look at it separately. So we're going to look at the interval of sequent to the 4th of theta the theta. And this is going to be equal to. The integral of and here we'll break up sequent to the 4th of theta into 2 sequent squares of theta and multiplying them together. So we'll have sequent squared of theta and then that gets multiplied by sequence squared of theta which we're going to rewrite as the quantity of 1 plus tangent squared of theta in quantity d theta. So again, we'll multiply this out and separate this into two integrals, so this is going to be equal to the integral of squad of the, the theta. Plus the interval of squared of theta. Multiplied by tangent squared of the, the theta. Now for the last integral there, we're going to want to evaluate the integral using new substitution. So we'll set U equal to tangent of data, and that means that DU is going to be equal to. Sin squared the the theta. And so rewriting our integrals, this is going to be equal to the integral of sequence squared the the theta plus the integral of U2. Do you So, performing the integrations, you see that this is going to be equal to. When we integrate sequence square theta with respect to data, we get tangent of data. Then we'll have plus and we integrate U2d with respect to you, we get you cubed divided by 3. Then we'll have to have our integration constant of plus C. Now we want to substitute in for u tangent of theta, so that means that this is going to be equal to tangent of theta plus 1 divided by 3 multiplied by tangent, cubed of theta plus C. So now we can evaluate our original in. So we'll have the original integral, which is the integral of 24th divided by the quantity of 1 plus z squared and quantity dz, and that is going to be equal to 2 multiplied by the quantity, and here we'll substitute in our result that we have for the integral of sequent to the 4th theta d theta, and that is tangent of theta plus 1 divided by 3 multiplied by tangent cubed of theta. In quantity, then we'll have minus 4 multiplied by the integral of sequence square theta the data, which is when we do that integration, tangent of data. Then we'll have + 2. Multiplied by the integral of the data, which is just data when we evaluate that integral, and then we will combine all of our integration constants into a single integration constant of plus C at the end. So now we just need to collect like terms, so this is going to be equal to. For the first term, we'll have 2 tangent of data, we also have a -4 tangent of data, so we'll have -2 tangent of data. Then we'll have plus 2 divided by 3 multiplied by tangent cute of data. Plus 2 theta plus Z. Now we want our final answer to be in terms of Z, and so if we look at our original definition of Z, we see that Z is equal to tangent of theta. Which means that data is going to be equal to the inverse tangent of Z. So, using these two results, we'll see that our integral is going to be equal to. Minus 2 Z plus 2 divided by 3, multiplied by Z cubed. Plus 2 multiplied by the inverse tangent of Z. Plus C. So 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.

7-56. Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.
54. ∫ y⁴/(1 + y²) dy

Key Concepts

Trigonometric Substitution

Integral of Rational Functions

Differentiation of Trigonometric Functions

Watch next