Evaluating $\int_0^{\pi/2} \frac{x^2}{\sin(x)} dx$

Asked Sep 06 '21 at 20:26

Active Sep 06 '21 at 20:37

Viewed 72 times

I have been trying to do the following integral: $$\int_0^{\pi/2} \frac{x^2}{\sin(x)} dx$$

My Attempt

$$\int_0^{\pi/2} \frac{x^2}{\sin(x)} dx= \int_0^{\pi/2} x^2\csc(x) dx= \Bigg[2x \csc(x)dx\Bigg]_0^{\pi/2} -\int_0^{\pi/2} 2x\int\csc(x)dx dx$$

$$\Bigg[2x \csc(x)\Bigg]_0^{\pi/2} -\int_0^{\pi/2} 2x\ln(\csc(x)-\cot(x)) dx $$ How do you proceed from here? How do you solve this integral? Thank you for your time.

edited Sep 06 '21 at 20:32

asked Sep 06 '21 at 20:26

hwood87

1,352

1

I think the integral does not converge as you wrote it. – Atmos Sep 06 '21 at 20:28
1

What about the points with $\sin x=0$? – user Sep 06 '21 at 20:29
2

This is clearly divergent. If you are interested in a continuation or some function that can be expressed in some region as an integral you might say so, but as-is there is no meaningful interpretation. – Integrand Sep 06 '21 at 20:29
The integral does not converge. – Zanzag Sep 06 '21 at 20:32
I corrected the bounds – hwood87 Sep 06 '21 at 20:33
1

Does this answer the question? – Paul Enta Sep 06 '21 at 20:38
I thought you were looking simply for a convergence test! – user Sep 06 '21 at 20:39
How do you explain the steps you did in integration by parts? – Laxmi Narayan Bhandari Sep 07 '21 at 03:46

Evaluating $\int_0^{\pi/2} \frac{x^2}{\sin(x)} dx$

0 Answers0