Please solve : ∫[pi,0] [sinx/(sinx+cosx)]dx

Please solve : ∫[pi,0] [sinx/(sinx+cosx)]dx?

[I think the pi would be π/2]

anyway...

Let.: I= ∫[π/2,0]

*[sin(x)/(sin(x)+cos(x)]dx...[i]

= ∫[π/2,0]
*[sin(π/2-x)/(sin(π/2-x)+cos(π/2-x)]dx

= ∫[π/2,0][cos(x)/(cos(x)+sin(x))]dx....[i…


now, [i]+[ii]

=> 2I= ∫[π/2,0][sin(x)/(sin(x)+cos(x)]dx+

∫[π/2,0][cos(x)/(cos(x)+sin(x))]dx


=>2I=∫[π/2,0]
*[(sin(x)+cos(x))/(sin(x)+cos(x))]dx

=> 2l= ∫[π/2,0][1]dx

=[x][π/2,0]

= [π/2-0]

=π/2

=>I=π/4

Can u give me some useful links for CBSE class XII math bord exam ?

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I am not sir ,but a little kid

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Let I = ∫(x = 0 to π) sin x dx/(sin x + cos x).

Let w = π - x, dw = -dx.

Since sin(π - w) = sin π cos w - cos w sin π = -cos w,
and cos(π - w) = cos π cos w + sin π sin w = -sin w,

I = ∫(w = π to 0) -cos w * -dw/(-cos w - sin w)
..= ∫(w = 0 to π) -cos w dw/(-cos w - sin w)
..= ∫(w = 0 to π) cos w dw/(cos w + sin w)
..= ∫(x = 0 to π) cos x dx/(cos x + sin x), dummy variable change.

Now, add this to the original version of I:
2I = ∫(x = 0 to π) 1 dx = π.
==> I = π/2.

I hope this helps!