# Integration by substitutionWatch

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#1
This one is pretty crazy, i think i got going but now i'm stuck...

Integrate √(a²-x²) dx with limits ½a to a

and show that it equals a²/2(π/3 - √3/4)

Thanks...

BTW this π is supposed to be a pi sign but it hasn't come out too well :-/
0
13 years ago
#2
set x = asin(x)
0
13 years ago
#3
try x = asint or something like that , inside the square root you will get a2cos2t , which is pretty funky
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13 years ago
#4
(Original post by Clearvision)
This one is pretty crazy, i think i got going but now i'm stuck...

Integrate √(a²-x²) dx with limits ½a to a

and show that it equals a²/2(π/3 - √3/4)

Thanks...

BTW this π is supposed to be a pi sign but it hasn't come out too well :-/
x = asint => dx/dt = acost => dx = acost.dt
=> rt(a^2-a^2sin^2t)
= rt(a^2cos^2t)
= acost

INT(a^2.cos^2t)dt
= a^2.INT(0.5+0.5cos2t).dt
= a^2(0.5t + 0.25sin2t) between x = 0.5a and x = a

x = 0.5a
=> asint = 0.5a
=> sint = 0.5
=> t = pi/6

x = a
=> asint = a
=> sint = 1
=> t = pi/2

INTEGRAL between stated limits
= a^2(0.5.pi/2 + 0.25sinpi) - a^2(0.5pi/6 + 0.25sinpi/3)
= a^2(pi/4+0) - a^2(pi/12+rt3/8)
= a^2/2(pi/2 - pi/6 - rt3/4)
= a^2/2(pi/3 - rt3/4)
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#5
Yeah i was using the substitution x=a sinθ

So dx = a cosθ + sinθ

but then i seem to be left with a mad integral with 2 variabes, θ and a.
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#6
Thanks widowmaker...but when you differentiate acost why don't you use the product rule to get acost + sint?
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13 years ago
#7
a is a constant, not a variable
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#8
Oh right, i see now.

Thanks
0
13 years ago
#9
(Original post by Clearvision)
Thanks widowmaker...but when you differentiate acost why don't you use the product rule to get acost + sint?
a is a constant, so there is no product of functions.
0
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