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
[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
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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!
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!