Find the indefinite integral of 5secxtanx dx. Consider restricted values of x.

[MathMajor123]

Hi, I am having trouble with this problem. I've made it to the expression:
5x sec(x) − 5ln(|sec(x) + tan(x)|) + C. How do I find the equivalent expression that accounts for restricted values of x, however?

Thanks for your help.

 

[Dr.Peterson]

Hi, I am having trouble with this problem. I've made it to the expression:
5x sec(x) − 5ln(|sec(x) + tan(x)|) + C. How do I find the equivalent expression that accounts for restricted values of x, however?

Thanks for your help.

Please show your work; the restriction may require a modification at some step, rather than something to be added at the end. Also, if you didn't show us the entire problem (including any choices, or a provided solution that you are comparing yours to) please do so; an image may be best. (I get a very different answer, too, so there may be an error in your work in the first place.)

If you were taught something about "restricted values" in an integral, be sure to read that, and perhaps even show us what was taught, so we can be sure what they have in mind.

 

[BigBeachBanana]

Hi, I am having trouble with this problem. I've made it to the expression:
5x sec(x) − 5ln(|sec(x) + tan(x)|) + C. How do I find the equivalent expression that accounts for restricted values of x, however?

Thanks for your help.

I got a different answer after integrating the function. The answer you provided does not vary by constant integration compared to mine. I've checked my answer with an engine, and it agrees. You may have made a mistake somewhere in your work.

The "restricted value of x" could be another way of asking for the function's domain because it does not exist everywhere.

 

[pka]

The title of your OP was: Find the indefinite integral of 5secxtanx dx

If you note that [imath]\sec(x)\tan(x)=\dfrac{\sin(x)}{\cos^2(x)}[/imath].
This implies [imath]\displaystyle\int {-\dfrac{{1}}{{{u^2}}}du} [/imath]

 

[MathMajor123]

My mistake, it seems I typed it incorrectly, the question asked for the indefinite integral of 5xsecxtanx dx. The correct answer is shown as being [math]5\left(x\sec \left(x\right)+\ln \left(\left|\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)\right|\right)-\ln \left(\left|\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)\right|\right)\right)+C[/math] . I have only reached 5x sec(x) − 5ln(|sec(x) + tan(x)|) + C by integrating by parts. How do I go from this expression to the final solution?

 

[Dr.Peterson]

My mistake, it seems I typed it incorrectly, the question asked for the indefinite integral of 5xsecxtanx dx. The correct answer is shown as being [math]5\left(x\sec \left(x\right)+\ln \left(\left|\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)\right|\right)-\ln \left(\left|\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)\right|\right)\right)+C[/math] . I have only reached 5x sec(x) − 5ln(|sec(x) + tan(x)|) + C by integrating by parts. How do I go from this expression to the final solution?

I can see that they are equal by graphing both on Desmos (yours in red, theirs in blue dots):



Since your form is much simpler than theirs, I see no reason to make it look like theirs. Rather, you might try simplifying theirs, using half-angle formulas, and hope it will look like yours. (Forms like this, however, can be rather slippery, and you may get a form that looks different, but is still equivalent.)

 

[BigBeachBanana]

I agree with Dr.Peterson that your answer is such simpler, but if you insist on matching the solution then (ignoring the log for now):
[math] \tan(x)+\sec(x)= \csc\left(\frac{\pi}{4}-\frac{x}{2}\right)\sin\left(\frac{x}{2}+\frac{\pi}{4}\right)\\ =\frac{\cancel{\sqrt{2}}}{\cos(\frac{x}{2})-\sin(\frac{x}{2})} \cdot\frac{\sin(\frac{x}{2})+\cos(\frac{x}{2})}{\cancel{\sqrt{2}}} [/math]Apply the log, and use the quotient rule, you'll get their answer.
[math]\ln\left|\frac{\sin(\frac{x}{2})+\cos(\frac{x}{2})}{\cos(\frac{x}{2})-\sin(\frac{x}{2})}\right|[/math]