Use the substitution z = tan(θ/2) to evaluate the integrals in...

Question

Use the substitution z = tan(θ/2) to evaluate the integrals in Exercises 41 and 42.

∫ csc θ 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. Use substitution. Z is equal to tangent of theta divided by 2 to evaluate the interval of sant of theta d theta. OK, so it appears for this particular problem, we're asked to use the substitution method on this equation for Z in order to evaluate this integral that is provided to us by the prom itself. So we need to use substitution considering our equation for Z in order to help us evaluate the integral that is provided to us by the prom itself. So with that in mind. Let us note that we need to use the substitution. Z is equal to tangent of theta divided by 2 in order to convert the integrand into a rational function of z and DZ. And then we will need to integrate the resulting rational function by using partial fractions, and then we will later need to back substitute and rewrite the integral in its standard form. So that was a quick run-through of what we're essentially going to do in order to solve this problem. So with that in mind, our first step that we need to take is we need to find the derivative of Z equal to tangent of theta divided by 2. And we need to, once again, to reiterate, we are taking the derivative of Z is equal to tangent of theta divided by 2. And when we take the derivative, we will find that DZ. Is equal to 1 divided by 2 multiplied by C 2 of theta divided by 2D theta. And our second step now is we need to change, let's make a note, we need to change C can't squared. Of theta divided by 2 in terms of the, fantastic. So we need to take C square of theta divided by 2, and we need to change it to terms of Z in terms in Z are terms of Z. So, with that said, we need to take D Z is equal to 1 divided by 2 multiplied by C square. Of theta divided by 2 d theta, which is equal to drum roll, 1 divided by 2 multiplied by square brackets of 1 plus tangent squared of theta divided by 2. D theta, which will be all equal to 1 divided by 2 multiplied by parentheses of 1 plus z to the 2 d theta. Our third step now is we need to solve for D Z is equal to 1 divided by 2 multiplied by parentheses of 1 plus z to the power of 2. And then we need to multiply this by DZ. So, we're gonna need to multiply this by DZ. Fantastic. Or specifically D theta. So once again, let's note that our third step is we need to solve for DZ is equal to 1 divided by 2, multiplied by parentheses of 1 + z to the power of 2. D theta, and we're solving for Dheta specifically. So, when we do that, we will find that D theta is equal to 2 multiplied by DZ divided by 1 plus Z to the power of 2. So, moving right along here, our 4th step now is we need to express c of theta, so we need to express cant of theta in terms of Z. So we need to take sec of theta, which is equal to 1 divided by cosine of theta, which is equal to 1 plus tangent squared. Of theta divided by 2 divided by 1 minus tangent square of theta divided by 2. And we need to note that this will be all equal to 1 + Z 2 divided by 1 minus Z 2. So, our 5th step now is we need to form the integrand in terms of Z. So we need to take C of theta, D theta, which will be equal to 1 + Z 2 divided by 1 minus Z to the power of 2, and we need to multiply this by 2. Divided by 1 + Z to the power of 2, DZ, which will be equal to 2 divided by 1 minus Z to the power of 2 DZ. Our step number 6 is we need to split the fraction of the integrand using the partial fractions. So using that methodology, we'll be able to write that the integral of 2 divided by 1 minus z to the power of 2 Dz will be equal to the integral of parentheses of 1 divided by 1 minus z. Plus 1 divided by 1 plus Z in parentheses DZ. Fantastic. Our step number 7 is we need to now integrate by recalling and using logarithmthic rules, so we need to use logs. And we'll be able to write the integral of parentheses of 1 minus, or rather, the integral of parentheses of 1 divided by 1 minus Z plus 1 divided by 1 + Z. Fantastic. In parentheses once again, and then we need to multiply this by DZ, which will be equal to negative the natural log of the absolute value of 1 minus Z. Plus the natural log of the absolute value of 1 + Z. Plus C. Fantastic. Well, let's take a moment to note that C denotes some constant interval. So, our step number 8 is we need to combine our logarithms, and when we do, we'll be able to write the natural log of the absolute value of 1 + Z divided by 1 minus Z. Fantastic. So now, step number 9, which is our final step, is we need to substitute back in Z is equal to tangent of theta divided by 2, so we're substituting back in our value of Z. And when we do, we'll be able to write that the integral of c can of theta d theta will be equal to the natural log of the absolute value of 1. Plus Tangent. Of, so it's gonna be tangent of. Theta divided by 2, all divided by 1 minus tangent of theta divided by 2. Don't forget that it's in absolute value signs plus C, and this is our correct and final answer. Hooray, we did it. We solved for this problem. So going back to the top to recap what we've learned, we can safely say that our correct and final answer for the evaluated integral once again without a doubt will be the natural log of the. Absolute value of 1 plus tangent of theta divided by 2, all divided by 1 minus tangent of theta divided by 2 plus C. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Use the substitution z = tan(θ/2) to evaluate the integrals in Exercises 41 and 42.∫ csc θ dθ

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Use the substitution z = tan(θ/2) to evaluate the integrals in Exercises 41 and 42.
∫ csc θ dθ