Evaluate the integrals in Exercises 39–54.
∫ 1 / (cos θ ...

Question

Evaluate the integrals in Exercises 39–54.

∫ 1 / (cos θ + sin 2θ) dθ

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. Evaluate the integral, the integral of 1 divided by sin of theta plus sine of 2 theta d theta. OK, so it appears for this particular prong we're simply asked to evaluate the integral that is provided to us. So now that we know that we're ultimately trying to evaluate this specific integral. Our first step that we need to take is we will need to simplify our integral by recalling and using double angle identities, so we need to first use the identity that states that. Sin of 2 theta, we need to recall and note that sin of 2 theta is also equal to 2 multiplied by sin of theta. Multiplied by cosine of theta, and we will need to use this identity in order to rewrite our denominator. So you can go ahead and write that the integral of 1 divided by sine of theta plus 2 multiplied by sine of theta multiplied by cosine of theta. D theta will be equal to the integral of 1 divided by sin theta multiplied by parentheses of 1 + 2 multiplied by cosine of theta. D theta, fantastic. So with that in mind, our second step now is we need to prepare to use U substitution. And in order to use U is equal to cosine of theta, we will first need to use, or rather, in order to use our u is equal to cosine of theta, we will first need to have a sin of theta located in the numerator, and we can achieve this by multiplying the top and bottom of our expression by sin of theta. So when we do that, we'll be able to write that the integral of sign of theta. Will be Divided by sin squared. Of theta multiplied by parentheses of 1 + 2 multiplied by cosine theta. fantastic d theta. So now, with that in mind, our next step now. We will now need to recall and use Pythagorean identity, which states that, let's make a note that states that sin squared of theta is equal to 1 minus cosine square of theta, and when we apply this identity, we can rewrite our integral as the integral of. Sine of theta divided by parentheses of 1 minus cosine squared of theta. Multiplied by parentheses of 1 + 2 multiplied by cosine of theta d theta. Fantastic. So now, for our third step is we need to perform our substitution. So let us state that you will be equal to cosine of theta. Which will mean that DU by D theta will be equal to negative sin theta, which will mean, therefore, that DU will be equal to sin theta d theta. Fantastic. So that will mean that we can write that the integral of sin of theta. Divided by parentheses of 1 minus. Oh sign Square of theta, so cosine squared of theta multiplied by parentheses of 1 + 2 multiplied by cosine of theta. D theta will be equal to the integral of -1 divided by parentheses of 1 minus u 2 multiplied by parentheses of 1 + 2 u. Do you Which will be equal to the integral of -1 divided by, so 1 divided by 1 minus u multiplied by 1 + u multiplied by parentheses of 1 + 2 u. Do you? OK. So, at this point, we now need to recall and use partial fraction decomposition. So we will now decompose the integrand into simpler fractions to help us solve for this problem in a more easy fashion. So we will take 1 divided by parentheses of 1 minusu multiplied by parentheses of 1 + U multiplied by 1 + 2U. Which will be equal to A divided by 1 minus U plus B divided by 1 + U, plus C divided by 1 + 2 U. And now we need to solve for our constants. So solving for A, we need to let you be equal to 1, which will mean that A is equal to -1 divided by. 2 multiplied by. 3, which will be equal to -16. Similarly, we need to solve for B and C now, so let's move on to B. So B, we need to let U be equal to -1, which will mean that B is equal to -1 divided by 2 multiplied by -1. Which will be equal to 1/2, and for c we need to let you be equal to 1/22, which will mean that c is equal to -1 divided by parentheses of 3 divided by 2 multiplied by parentheses 1/2. Which will be equal to -4 divided by 3. Thus, we can go ahead and write that 1 divided by 1 minusu multiplied by 1 + u multiplied by parentheses 1 + 2U will be equal to -1 divided by 6 multiplied by 1 minusu. Plus 1 divided by 2 multiplied by 1 plus you. -4 divided by 3 multiplied by 1 + 2 u in parentheses. So now, that will mean that our integral of -1 divided by 1 minus U multiplied by 1 + U multiplied by 1 + 2U. Do you? Fantastic. Will be once again now equal to drum roll, the integral of parentheses of -1 divided by 6 multiplied by 1 plus u. Plus 1 divided by 2 multiplied by 1 plus you. -4 divided by 3 multiplied by 1 + 2 U. Once again in parentheses, do you. OK, so now for our 5th and final step is we need to perform our integration and we also need to back substitute. And in order to do that, we need to recall and use the following formula which states that the interval of 1 divided by 1 plus a multiplied by u. DU will be equal to, so this will be equal to 1 divided by a multiplied by the natural log of the absolute value of 1 plus a multiplied by u plus c where c denotes some constant variable, so that will mean we can go ahead and write that the integral of parentheses of. -1 divided by 6 multiplied by 1 minus you. Plus 1 divided by 2 multiplied by 1 + U minus 4 divided by 3 multiplied by 1 + 2U. DU will be all equal to 1 divided by 6 multiplied by the natural log of parentheses 1 minus u. Plus. 1/2 multiplied by the natural log of 1 + u minus 2 divided by 3 multiplied by the natural log of the absolute value of 1 + 2 u. Plus C, fantastic. So moving right along here, we now need to substitute back u as equal to cosine and theta into our expression. And when we do that, we'll be able to write that 1 divided by 6 multiplied by the natural log of parentheses of 1 minus u + 1/2 multiplied by the natural log of 1 + u. Minus 2 divided by 3 multiplied by the natural log of the absolute value of 1 + 2 u plus c will be all equal to 1 divided by 6 multiplied by the natural log of parentheses of 1 minus cosine theta. Plus 1 divided by 2 multiplied by the natural log of parentheses of 1 plus cosine of theta. -2 divided by 3 multiplied by the natural log of the absolute value of 1 + 2 multiplied by cosine of theta plus c and that is our final answer. Hooray, we did it. We saw for our final answer. So going back to look at the prom itself, we could safely say that once again our final answer for this evaluated integral is as we should know by now. At once again is. 1/6 or 1 divided by 6 multiplied by the natural log of parentheses 1 minus cosine theta plus 1/2 multiplied by the natural log of parentheses of 1 plus cosine theta minus 2 divided by 3 multiplied by the natural log of the absolute value of 1 + 2 multiplied by cosine of theta plus c. And that's it. 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 39–54.∫ 1 / (cos θ + sin 2θ) dθ

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Evaluate the integrals in Exercises 39–54.
∫ 1 / (cos θ + sin 2θ) dθ