Integrating $\frac{x}{1 + \sin(x) \cos(x)}$ from 0 to $\pi$

  • Thread starter Thread starter kelvin911
  • Start date Start date
  • Tags Tags
    Integrating
kelvin911
Messages
2
Reaction score
0
<br /> \int_0^{\pi} \frac{x}{1+\sin(x)\cos(x)} dx<br />
 
Last edited:
Physics news on Phys.org
how does latex in this forum works? how to get rid of the 18416 ?

testing

\frac{1}{2}

\int_0^{\pi} \frac{x}{1+\sin(x)\cos(x)} dx
 
Why is this a hard integral? Post your work please
 
what is the method used?
 
This is not a competition. I had the answer the minute I looked at the problem to make sure it existed. The point is to teach him how to find the solution by looking at his attempt at the solution. Please don't randomly post an answer which does not help him in any way.


To OP: for what its worth, sin(2*x) = 2*sin(x)*cos(x). You can try u=2*x, du = 2 dx
 
Last edited:
cronxeh said:
This is not a competition. I had the answer the minute I looked at the problem to make sure it existed. The point is to teach him how to find the solution by looking at his attempt at the solution. Please don't randomly post an answer which does not help him in any way.

Exactly. I've deleted the posts with answers in them. This is most likely homework, so giving the OP the answer is cheating. I'm moving this thread to Homework Help, Calculus and Beyond.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top