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Integration (1 Viewer)
- Thread starter Faera
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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.
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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Sorry, I'm not following you, can you be more specific?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
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.
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.
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.
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.
freaking_out
Saddam's new life
oiii, me wanna cake too.
Grey Council
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lol, this is what happens when girls do 4u maths.
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Giant Lobster
Active Member
lmao
girls doing 4u math... theres technically nothing wrong with that
Grey Council
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technically...
but with cakes being passed around, i dunno. ^_^
ANYWAY:
let Faera do her maths. no more spam.
but with cakes being passed around, i dunno. ^_^
ANYWAY:
let Faera do her maths. no more spam.
freaking_out
Saddam's new life
leave me alone all of u mean ppl... i hardly spam,
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Grey Council
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ha! you greedy greedy people.
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