integration

[alexa_xox]

I need help finding the integral of (x/sinx + cosx) dx , between pi/2 and 0.
My teacher gave us the answer : pi/2root2 ln((root2) + 1)
I have no idea how to start working on this...if anyone could help in anyway at all, even if it's just the starting it off, i would greatly appreciate it!
thanks,
xox

 

[Deleted member 4993]

I need help finding the integral of (x/sinx + cosx) dx , between pi/2 and 0.
My teacher gave us the answer : pi/2root2 ln((root2) + 1)
I have no idea how to start working on this...if anyone could help in anyway at all, even if it's just the starting it off, i would greatly appreciate it!
thanks,
xox

Is your problem

\(\displaystyle \int_0^{\frac{\pi}{2}}[\frac{x}{sin(x)} \, + \, cos(x)] dx\)

or

\(\displaystyle \int_0^{\frac{\pi}{2}}[\frac{x}{sin(x) \, + \, cos(x)}] dx\)

or something else...

 

[fasteddie65]

If your problem is the second version, i.e. x/(sin x + cos x), then you might try multiplying both numerator and denominator by sin x - cos x. That is a useful technique in many situations. It just might help here.

 

[alexa_xox]

It's the integral of x/(sinx +cosx) * dx
The sinx and cosx all in the denominator.
I've tried using t-substitution but that doesn't work, and I've also tried multiplying by (sinx - cosx) but failed to get it out.
I'll keep trying!
thanks,
xox

 

[Deleted member 4993]

It's the integral of x/(sinx +cosx) * dx
The sinx and cosx all in the denominator.
I've tried using t-substitution but that doesn't work, and I've also tried multiplying by (sinx - cosx) but failed to get it out.
I'll keep trying!
thanks,
xox

\(\displaystyle \sin(x) \, + \, \cos(x) \, = \, \sqrt{2}\cdot \cos(x \, - \, \frac{\pi}{4})\)

transform

u = x - ?/4

and use the fact that x/cos(x) is an odd-function.

 

[alexa_xox]

thank you...also for anyone else whose reading i've figured you can also use ?a f(x)dx = ?a f(a-x)dx and then use t-substitution or what was advised previously, using harmonic form.
thanks