76–83. Preliminary steps The following integrals require a pre...

Question

76–83. Preliminary steps The following integrals require a preliminary step such as a change of variables before using the method of partial fractions. Evaluate these integrals.

79. ∫ [sec t / (1 + sin t)] dt

Accepted Answer

Hello there. Today we're 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 interval of Kican of X divided by 1 plus cosine of XDX. Awesome. So it appears for this particular problem we're ultimately trying to evaluate this specific interval. So now that we know that we're trying to evaluate this interval, our first step that we need to take in order to solve this problem is we need to recall and use the substitution method. So we need to substitute that U is equal to cosine of x. That would then that would therefore mean. That DU is equal to negative sign of X D X. So that will mean that DX is equal to negative DU divided by of X. So that will mean that the integrand will become 1 divided by sin of x multiplied by parentheses 1 plus cosine of X D X is equal to 1 divided by of X multiplied by parentheses 1 plus U. Multiplied by parenthesis negative DU divided by sin of X, which will be equal to negative, so it's gonna be equal to negative. DU divided by sin squared of X multiplied by 1 plus U. Fantastic. So, however, but we need to note, here's a big but that sin squared of X is equal to 1 minus cosine squad of X, which is equal to 1 minus u to the 2. So therefore we could write that the integral of cosicant of X divided by 1 plus cosine of X D X is equal to negative, the integral of DU divided by parentheses 1 minus U2 in parentheses multiplied by 1 + u. In parentheses. Which will be equal to negative, the interval of DU divided by 1 minus U multiplied by 1 + U to the power of 2. Fantastic. So now we must evaluate, so we need to take negative, the interval of DU divided by parentheses 1 minus U multiplied by parentheses 1 plus U to the power 2. Fantastic. Our second step now is we need to recall and apply partial fraction decomposition, so we need to write that 1 divided by parentheses 1 minus u multiplied by parentheses 1 plus U 2 is equal to a divided by 1 minus u plus B divided by 1 plus U. Plus C divided by parentheses 1 plus U to the power of 2. Awesome. So now, we need to multiply through by parentheses 1 minus U, multiplied by 1 plus U in parentheses to the 2. And when we multiply through by this expression, we will find that 1 is equal to a multiplied by parentheses 1 plus U to the power of 2, plus B multiplied by parentheses 1 minus U multiplied by parentheses 1 plus U, plus C multiplied by parenthesis of 1 minus U. Fantastic. So, moving right along here. We now need to expand and collect the powers of you, so we need to take 1 is equal to a multiplied by parentheses 1 + 2 U plusu to the power of 2. In parentheses plus B multiplied by parentheses 1 minus U 2 plus C multiplied by parentheses 1 minus U. Fantastic. So now we need to group the coefficients, and that will give us the system. So grouping coefficients gives the system, so we need to take the curly bracket of u to the power of 2, which is a minus B is equal to 0. You will give us 2A minus C is equal to 0, and the constant, which I'll just call C for short, will be A. Plus B + C will be equal to 1. So we need to note that A minus B will give us 0, we will get B is equal to A and from 2A minus C is equal to 0, we will find that C is equal to 2A, and we can substitute this into our constant equation which will give us a plus A plus 2 A is equal to 1, which will give us 4 A is equal to 1, which will give us an answer that A is equal to 1/4. Thus, B will be equal to 1/4 and C will be equal to 1/2. So, we can go ahead and write that 1 divided by parentheses 1 minus U, multiplied by parentheses 1 plus U 2 will be equal to 1/4 multiplied by 1 divided by 1 minusu. Plus 14th multiplied by 1 divided by 1 + U, plus 1/2 multiplied by 1 divided by parentheses 1 plus U to the power of 2. Fantastic. Our third step now is we need to integrate, so we need to recall that the interval has an overall minus sign, so we could say that the interval of Posicant of X, so cosicant of X divided by One plus cosine of XDX. is equal to negative the interval of 1/4 multiplied by 1 divided by 1 minus u plus 1/4 multiplied by 1 divided by 1 plus U, plus 1/2 multiplied by 1 divided by parentheses 1 + U to the power of 2. Do you Fantastic. And now we need to integrate term wise, so we need to take the integral of, so we need to take the integral of DU divided by 1 minus U, which will be equal to negative, the natural log of the absolute value of 1 minus U. And we need to take the integral of DU divided by 1 plus U, which will be equal to the natural log of the absolute value of 1 plus U. Fantastic. And finally, we need to take the interval of DU divided by parentheses 1 plus U to the power of 2, which will be equal to -1 divided by 1 plus U. Fantastic. So, with that in mind, we can go ahead and write that the interval of cosicant of X divided by 1 plus cosine of X D X will be equal to negative parentheses 14th multiplied by negative, the natural log of the absolute value of 1 minus U. Plus 1/4 multiplied by the natural log of the absolute value of 1 plus you. Plus 1/2 multiplied by parenthesis, negative 1 divided by 1 plus U. Plus C, which will be all equal to 1/4 multiplied by the natural log of the absolute value of 1 minus U. Minus 1/4 multiplied by the natural log of the absolute value of 1 + U plus 1 divided by 2 multiplied by parentheses 1 plus U plus C where C denotes some constant value. Fantastic. We now need to combine logs, and when we combine logs, we will find that 1/4 multiplied by the natural log of the absolute value of 1 minus cosine of x divided by 1 plus cosine of X plus 1 divided by 2 multiplied by parentheses 1 plus cosine of X. Fantastic. Plus C Fantastic. And let us note that we will let T equal tangent of X divided by 2. And because we're letting T equal tangent of X divided by 2, we can now recall and use the half-angle identities, which will state that Cosine of X is equal to 1 minus T 2 divided by 1 + T 2. And that 1 minus cosine of X is equal to 2 multiplied by T to the power of 2, divided by 1 plus T to the 2. And that 1 plus cosine of X will be equal to 2 divided by 1 + T to the power of 2. So, with that in mind, we will get that. Drum roll that 1 minus cosine of x divided by 1 plus cosine of X will be equal to drum roll. 2 multiplied by T 2 divided by 1 + T 2 divided by 2 divided by. 1 plus T to the power of 2, which will be simply equal to T the power of 2. So with that in mind, moving right along here, therefore, without a doubt, we could go ahead and write that 14th multiplied by the natural log of the absolute value of 1 minus cosine of X divided by 1 plus cosine of X will be equal to. 1/4 multiplied by the natural log of the absolute value of T to the power of 2, which will be equal to 12 multiplied by the natural log of the absolute value of T. Also, we can go ahead and write that 1 divided by 2 multiplied by the parentheses 1 plus cosine of X will be equal to 1 divided by 2 multiplied by 2 divided by 1 + T 2, which will be equal to 1 + T 2 divided by 4. Fantastic. We are on a roll here. So with that in mind, the whole expression will become the following. So we can go ahead and write that one half multiplied by the natural log of the absolute value of tangent of X divided by 2 + 1 + T to the power 2 divided by 4 + C. Or when we absorb the constant 1/4 or 1 divided by 4 into our constant C, we can go ahead and write our final answer as such 2 multiplied by the natural log of the absolute value of tangent of X divided by 2. Plus, Tangent squared of X divided by 2. Divided by 4. Plus C, and that without a doubt is our final answer. Hooray, we did it. So looking at the prom itself to recap what we've learned, so that will mean once again that our final answer, without a doubt will have to be drumroll, looking at our problem here. We have to be once again. 2, so we're gonna take 2 multiplied by the natural log of the absolute value of tangent of X divided by 2, plus tangent squared of X divided by 2. All divided by 4 plus C, where once again C denotes some constant value, 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.

76–83. Preliminary steps The following integrals require a preliminary step such as a change of variables before using the method of partial fractions. Evaluate these integrals.79. ∫ [sec t / (1 + sin t)] dt

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76–83. Preliminary steps The following integrals require a preliminary step such as a change of variables before using the method of partial fractions. Evaluate these integrals.
79. ∫ [sec t / (1 + sin t)] dt