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Integration (1 Viewer)

Faera

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Hey, i'm stuck on a few integration questions, any help would be very nice... :)

Find the indefinite integrals of:

1. (x^2 + 1)^(1/2)

2. cotx / [ln(esinx)]

3. [1/ (2x - x^2)^1/2]


Thanks in advance!
 

CM_Tutor

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Integral of (x<sup>2</sup> + 1)<sup>1/2</sup> is a substitution problem, let x = tan@

Integral of cot x / ln(esin x) - an interesting problem. Use log laws to rewrite the denominator as 1 - ln(cosec x), then use the substitution u = cosec x to transform the problem to int 1 / [u(ln u - 1) du. A second substitution, v = ln u, gives int 1 / (v - 1) dv. Thus, the answer is ln |v - 1| + C = ln |ln u - 1| + C = ln |ln(sin x) + 1| + C

Integral of 1 / (2x - x<sup>2</sup>)<sup>1/2</sup>, complete the square on x, and rewrite as 1 / sqrt[1 - (x - 1)<sup>2</sup>]. It integrates to an inverse trig function.
 
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Faera

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oooh- okay- thanks. :)
One thing- with the (x^2 + 1)^(1/2) problem, my tutor said something about substituting (x^2 + 1)^(1/2) = u - x which i dont really understand; he said it'd work if I managed to make the x^2's cancel out... 0.o
 

CM_Tutor

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Originally posted by Faera
oooh- okay- thanks. :)
One thing- with the (x^2 + 1)^(1/2) problem, my tutor said something about substituting (x^2 + 1)^(1/2) = u - x which i dont really understand; he said it'd work if I managed to make the x^2's cancel out... 0.o
Sorry, I'm not following you, can you be more specific?
 

Faera

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OK- for the integral of (x^2 + 1)^(1/2),
You said to let x = tan@, correct?
My tutor told me to try the problem by substituting:
u - x = (x^2 + 1)^(1/2), then squaring both sides to make x^2 disappear.
Which I've tried to do- but it doesn't seem to work in actually getting the integral.
 

CM_Tutor

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OK, when you set u - x = sqrt(x<sup>2</sup> + 1), you can rewrite it so that u<sup>2</sup> = 1 + 2ux ___(*), and so dx = (u - x)du / u. As a result, we have int (u - x)<sup>2</sup>du / u. Making x the subject of (*) and substituting gives int (u<sup>2</sup> + 1)<sup>2</sup>du / 4u<sup>3</sup>, which can be integrated.

This would be a difficult trick to see, and I'm not sure that it actually makes life that much easier in this case, so make sure you can do the standard approach as well. :)
 

Grey Council

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lol, this is what happens when girls do 4u maths.
 
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shkspeare

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i rkn.. he never posts nething constructive...

its either "LOL HAHAHA"
or
"hehe yeah!"

well not exactly that.. but spam neway

no offense :)
 

Grey Council

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technically...
;)

but with cakes being passed around, i dunno. ^_^

ANYWAY:
let Faera do her maths. no more spam.
 

CM_Tutor

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Hey, everyone, leave Faera alone. You're all just jealous that you didn't get the cake! :p
 

Faera

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Haha... awwww....

*gives you some lamingtons to make up for the stolen cakes*

geez... this thread has become so .... cake-oriented...
 

McLake

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Slighlty off topic ....

(I'll let is slide if someone offers me some nice cake)
 

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