#Definite integral
jade estuary
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jade estuary
im getting 2pi but i have been on drugs lately someone pls confirm
,w integrate (5sin(x)+3cos(x))/(sin(x)+cos(x)) from 0 to pi/2
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jade estuary
i trust your process
jade estuary
i trust your process
ah thanks
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@serene fossil has given 1 rep to @worldly hare
jade estuary
,w integrate x^4/[(x-1)(x^2+1)]
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jade estuary
,w integrate x*sin^-1x
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swift igloo
??
jade estuary
,w integrate xtanx /(secx*cosecx) from 0 to pi
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jade estuary
,w integrate (x^2-3x+1)/(1-x^2)^1/2
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jade estuary
,w integraet x^2/(1-x^2)^1/2
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jade estuary
,w convert sin^x in terms of tan^-1
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Requested by ._corazon._.
jade estuary
,w sin^-1 x = tan^-1 y
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jade estuary
,w differentiate x/(1-x^2)^1/2 -1
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upper pasture
stable egret
Let's look at a slightly more general case.
(a cos(x) + b sin(x))/(cos(x) + sin(x)) = (a cos(x) + b sin(x))(cos(x) - sin(x))/(cos(x)^2 - sin(x)^2) = (a cos(x)^2 + (b - a)sin(x)cos(x) - b sin(x)^2)/cos(2x) = (1/2)(a (1 + cos(2x)) + (b - a)sin(2x) - b (1 - cos(2x)))/cos(2x) = (1/2)(a - b + (b - a)sin(2x) + (a + b)cos(2x))/cos(2x)
Let's transform this a bit more.
(a cos(x) + b sin(x))/(cos(x) + sin(x)) = (1/2)(a + b) + (1/2)(a - b)(1 - sin(2x))/cos(2x)
That last term is cot(x + π/4). So:
(a cos(x) + b sin(x))/(cos(x) + sin(x)) = (1/2)(a + b) + (1/2)(a - b)cot(x + π/4)
And now it's easy, since cot(x + π/4) is odd around x = π/4.
normal wadi
Let's look at a slightly more general case.
(a cos(x) + b sin(x))/(cos(x) + sin(...
If you use the kings rule (or equivalently the substitution y=pi/2-x) it becomes quite simple (considering what the bounds of integration are)
stable egret
If you use the kings rule (or equivalently the substitution y=pi/2-x) it become...
Ah, yeah, that's true.
stable egret
Did u actually integrate it indefinitely
Yeah. Why not? 😄
jade estuary
Yeah. Why not? 😄
Caus no point literally!
stable egret
Caus no point literally!
Well, I was interested in the result.
That approach is longer, but it's not like it's invalid.
stable egret
I multiplied the numerator and denominator by (cos(x) - sin(x)), then simplified and applied some identities.
arctic stratus
if it was indefinite integral
Nr=A(d/dx Dr)+B(Dr)
works
Nr=numerator
Dr=denominator
it also works for $\frac{ae^x+be^{-x}}{ce^x+de^{-x}}$
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pratham
stable egret
if it was indefinite integral
Nr=A(d/dx Dr)+B(Dr)
works
Nr=numerator
Dr=denomin...
Oh, great idea! Didn't think of that.
stable egret
it also works for $\frac{ae^x+be^{-x}}{ce^x+de^{-x}}$
Oh, that's pretty much the same, then, as we can just rewrite everything in terms of complex exponents.
arctic stratus
Oh, great idea! Didn't think of that.
we were thought that under 'methods of integration'
different forms of integrals were classified
and they had a common approach
stable egret
I see. We were also taught in this way, but we didn't have this approach. Good to know!
jade estuary
if it was indefinite integral
Nr=A(d/dx Dr)+B(Dr)
works
Nr=numerator
Dr=denomin...
We are taught this for polynomials though