Show the two definite integrals are equal

[Im Kristin]

Hi guys. I need to show these two definite integrals are equal, but I have no idea Can you please have a look? Thanks!

 

[skeeter]

[imath]\dfrac{\cos{x}-\sin{x}}{1+\sin(2x)} = \dfrac{\cos{x}-\sin{x}}{\cos^2{x}+2\sin{x}\cos{x}+\sin^2{x}} = \dfrac{\cos{x}-\sin{x}}{(\cos{x}+\sin{x})^2}[/imath]

[imath]u = \cos{x} + \sin{x} \implies du = (\cos{x} - \sin{x}) \, dx[/imath]

perform the substitution ...

 

[Dr.Peterson]

Hi guys. I need to show these two definite integrals are equal, but I have no idea Can you please have a look? Thanks!
View attachment 32107

I suspect that the context of the problem would give some hints as to what method you are expected to try. It sounds like you aren't supposed to actually integrate anything, so it's probably some property you are to apply.

Very likely it is something about symmetry of some sort. For example, you might try replacing x in the first integral with pi/2 - x (which is a reflection) and see if you get the second integral, showing that they are equal. Is that something you have recently learned?

 

[Steven G]

[imath]\dfrac{\cos{x}-\sin{x}}{1+\sin(2x)} = \dfrac{\cos{x}-\sin{x}}{\cos^2{x}+2\sin{x}\cos{x}+\sin^2{x}} = \dfrac{\cos{x}-\sin{x}}{(\cos{x}+\sin{x})^2}[/imath]

[imath]u = \cos{x} + \sin{x} \implies du = (\cos{x} - \sin{x}) \, dx[/imath]

perform the substitution ...

Did you see that your two substitution would help? If yes, that is simply brilliant. I knew for a long while that you were amazing but this added on some more. Good job!