Intergration question (1 Viewer)

[stainmepink]

given that ∫f(x)dx [integral from a~0] = ∫f(a-x)dx [integral from a~0]

show the integral [from pi/2~0] of

cosx dx
---------------- = (pi/4)
cosx + sinx

help my test is tomorrow!

 

[_ShiFTy_]

I = ∫[pi/2-->0] (cosx)dx/(cosx + sinx)....(1)
= ∫[pi/2-->0] [ (cos (pi/2 - x)dx ] / [ cos(pi/2 - x) + sin(pi/2 - x) ]
= ∫[pi/2-->0] (sinx)dx/(sinx + cosx)....(2)

(1) + (2)

2I = ∫[pi/2-->0] (sinx + cosx)/(sinx + cosx)dx
= ∫[pi/2-->0] 1.dx
= x | [pi/2-->0]
= pi/2
I = pi/4

 

[pLuvia]

You can use t=tan(x/2) here as well, but shifty's way faster

 

[stainmepink]

thanks guys!!


i had my test today and that question came up, thanks to you i could do it

although i left it as 2I = ..

 

[_ShiFTy_]

You will lose one mark for not changing it back to 'I' from '2I'...unless the question specifically asked for 2I which i highly doubt