2

$$ \int_0^{\pi/2}\frac{\cos 3x+1}{\cos 2x-1}dx $$

Set $t=\sin x\implies dt=\cos x.dx$ $$ \int_0^{\pi/2}\frac{\cos 3x+1}{\cos 2x-1}dx=\int_0^{\pi/2}\frac{4\cos^3x-3\cos x+1}{-2\sin^2x}dx\\ =-2\int_0^{\pi/2}\frac{(1-\sin^2x)\cos x}{\sin^2x}dx+\frac{3}{2}\int_0^{\pi/2}\csc x\cot x.dx-\frac{1}{2}\int_0^{\pi/2}\csc^2x.dx\\ =-2\int_0^{1}[t^{-2}-1]dt+\frac{3}{2}\int_0^{\pi/2}\csc x\cot x.dx-\frac{1}{2}\int_0^{\pi/2}\csc^2x.dx\\ =-2\Big[\frac{-1}{t}-t\Big]^1_0-\frac{3}{2}\Big[\csc x\Big]_0^{\pi/2}+\frac{1}{2}\Big[\cot x\Big]_0^{\pi/2} $$

How do I apply the limits here as $t\to0\implies\frac{1}{t}\to\infty$, similarly for $\cot x, \csc x$ etc ?

Note: Solution given in my reference1 is $1$

Sooraj S
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  • The function is not continuous at zero (vertical asymptotic line), does the integral converges at all?! – K.K.McDonald Jan 10 '20 at 19:19
  • @ss1729 i upvoted. interesting question, with good work shown. what happens if you try to express the entire answer (i.e. all 3 integrals) in terms of $t$, rather than expressing 2 of the integrals in terms of $x$. i-m too lazy to try that myself. – user2661923 Jan 10 '20 at 19:20
  • Related https://math.stackexchange.com/questions/3428840/compute-int-0-pi-2-x2-left-sum-n-1-infty-1n-1-cosnx-cosnx – Ali Shadhar Jan 10 '20 at 19:25
  • @user2661923 in either case how do I deal with such situations where substituting the limits give infinity ? – Sooraj S Jan 10 '20 at 19:41
  • @ss1729 i was metacheating (i.e. hoping that the problem would not have been posed by your teacher if there wasn't a finite answer). my previous comment was intended to suggest an approach where all the bothersome terms might cancel out. – user2661923 Jan 10 '20 at 19:43
  • @K.K.McDonald My reference gives the solution 1, but not sure about it. – Sooraj S Jan 10 '20 at 20:02
  • @José Carlos Santos could you please explain how should we deal with such integrals when it becomes infinity whn the limits are applied ? – Sooraj S Jan 10 '20 at 20:15
  • @ss1729 Since it's not convergent (see my answer), it can't be 1. Perhaps you copied the problem wrong or something? – bjorn93 Jan 10 '20 at 23:48

2 Answers2

1

Hint

The numerator can be written as $$\cos3x+\cos x+1-\cos x=2\cos2x\cos x+1-\cos x=2\cos x(\cos2x-1)+1+\cos x$$

Now $$-\dfrac{1+\cos x}{1-\cos2x}=-\left(\dfrac{\cos\dfrac x2}{\sin x}\right)^2$$

Now $$\dfrac{\cos\dfrac x2}{\sin x}=\dfrac{\csc\dfrac x2}2$$

lab bhattacharjee
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  • how do I substitute the limits if it gives infinity ? – Sooraj S Jan 10 '20 at 19:38
  • here if you substitute the limits it gives infinity, right ?. My doubt is what do we do in such cases ? and how do we justify that ? – Sooraj S Jan 10 '20 at 19:47
  • @ss1729 If you substitute proper limits and get infinity then the answer is infinity! I'm not sure why you think that infinity is an invalid answer. That is the expected result since the series is non-convergent in nature. – Sam Jan 11 '20 at 04:02
0

The integral doesn't converge. Take its negative: $$\int_0^{\pi/2}\frac{\cos(3x)+1}{1-\cos(2x)}\,\mathrm{d}x $$ As $x\to 0^+$, $\cos(3x)+1\to 2$ and $1-\cos(2x)\sim 2x^2$, so $$\frac{\cos(3x)+1}{1-\cos(2x)}\sim\frac{1}{x^2}$$ and it's easy to verify that $$\int_0^{\pi/2}\frac{1}{x^2}\,\mathrm{d}x $$ is not convergent.

bjorn93
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