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09-01-2012, 20:08   #1
Iancar29
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Integration by parts help.

Hey people , need some help with this Question if anyone is willing.


Ok heres the sum ∫ Sin x ( 1+ √ (cosx) )^2 dx



So my guess is : u = Sinx and dv = ( 1+ √ (cosx) )^2 dx




Then ..... du = cosx v = ∫ ( 1+ √ (cosx) )^2 dx
*tried expanding it to... *
v= ∫ ( 1 +2√ (cosx ) + cos x ) dx
v = x + "?" + sinx

Its the v part i cant figure out ... is it a double integration by parts question?


Any help much appreciated...

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09-01-2012, 23:31   #2
RolandIRL
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I'd usually go with a substitution method for simple trig integrals (involving a mixture of sin and cos only) rather than integration by parts. Let u equal some part of the function and you'll usually end up with the derivative involving some term on its own that's outside the brackets.

Integration by parts (been a while since i've had to do it so might be mistaken) usually involves different types of functions like log and trig multiplied together or something that isn't defined in the log tables. Such examples would be something like integral of log(x) dx or integral of x*cos(x) dx.
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09-01-2012, 23:49   #3
ray giraffe
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It's integration by substitution u=cos(x)
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13-01-2012, 16:31   #4
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Thanks lads, im not so great using the subbing method on fractions havent done it in ages

So im doing loads of examples.

any help with this one please?


∫ ( 2x^3 + 10x ) / (x^2+1)^2 dx

how do i decide the u in this case?

I tried to expand the denominator to make it cancel with some of the numerator .But with no luck.

Once again any help much appreciated.

Last edited by Iancar29; 13-01-2012 at 16:37.
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13-01-2012, 18:49   #5
Iancar29
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Right so, with some searching myself i found it was a separation of partial fractions method.


Everything went alright except in the end when i come to filling everything back in i still end up with 2 not so nice integrals.

heres the 3 i ended up with.


∫ -4 / (x+1) + ∫ (5x+5) / (x-1) + ∫ 2x / ( X^2+2x+1)

-4Ln( x+1) + ? Ln(x-1) + ?


ANyone help to resolve the 2nd 2?
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13-01-2012, 19:39   #6
RolandIRL
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It's still a substitution question so use that with the original integral.

Generally for integrals involving something raised to a power in the denominator, you let u equal what's inside the bracket in the denominator. In this case let u equal (x^2 + 1), factorise the top part, and you'll find that du can be expressed involving some terms in the numerator.
From there, you'll end up with something like

integral of (u + c)/u^2 du

which should be easy to do.

If you're still having trouble grasping what method to use to solve integrals, look up some videos on Khan Academy. Linked the calculus section there. If you start from "The Indefinite Integrals or Anti-Derivative", that might help you understand integrals better
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13-01-2012, 23:33   #7
Colours
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The Integration By Parts Rule says that

So if we first simplify it a little by substituting cosx for z then the expression to be integrated becomes:

-∫(1+√ z)^2dz = -∫udv=-[uv-∫vdu] (When we have integrated this expression then we can replace z with cosx once again)

Now we apply Integration By Parts to this version taking u to be (1+√z)^2 and dv to be dz. We work out from this that du/dz= 2(1+√z)(1/√z)(1/2) which leads to du=(1/√2+1)dz.

And also much more straightforwardly v=z

So plugging both of these into our Integration By Parts formula gives:
-∫(1+√ z)^2dz= -[((1+√z)^2)(z)-∫(z)(1/√z+1)dz]

= -[z(1+√z)^2-∫(√z+z)dz]

=-[z(1+√z)^2-(2/3*z^(3/2)+(1/2z^2]

=-z(1+√z)^2+2/3z^(3/2)+1/2z^2

Multiplying out the square of the expression in brackets on the left:

= -z(1+2√z+z)+2/3z^(3/2)+1/2z^2

=-z-2z√z-z^2+2/3z^3/2+1/2z^2

=-z-2z^3/2-z^2+2/3z^3/2+1/2z^2

Since we've eliminated all of the integrals we can now substitute back
z for cosx which gives our final answer:

-cosx -2(cosx)^3/2-(cosx)^2+2/3(cosx)^3/2+1/2(cosx)^2

So I haven't yet figured out how to present maths expressions in a clearer way on here! Will look into it now.

Last edited by Colours; 13-01-2012 at 23:36.
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15-01-2012, 20:53   #8
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Thanks alot for the helP!
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13-05-2012, 22:24   #9
MathsArthur
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there is a much simpler solution to this. again use u=cos x, then du = -sin x dx and

INT (sin x (1 + cos^(1/2) x)^2)dx = -INT (1 + u^(1/2))^2 du

= -INT (1 + 2u^(1/2) + u) du = - (u + 4/3 u^(3/2) + u^2/2)

= -cos x - 4/3 cos^(3/2) x -cos^2 x/2 as before.
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13-05-2012, 22:55   #10
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for the integral

INT (2x^3 + 10x)/(x^2 + 1)^2 dx

= INT (2x^3 + 2x + 8x)/(x^2 + 1)^2 dx

= INT (4x(x^2+1)/2(x^2+1)^2 + 4*2x/(x^2 + 1)^2 dx

= (1/2)ln((x^2+1)^2) - 4/(x^2 + 1)

= ln(x^2+1) - 4/(x^2 + 1)
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