How to apply limits for the definite integral $\int_0^{\pi/2}\frac{\cos 3x+1}{\cos 2x-1}dx$

Question

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

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 $1$

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

Answer 2

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.