Using substitution to reduce integrals to standard form (1 Viewer)

WEMG · Jan 8, 2011

In chapter 5.3 in Cambridge 4u there is an example:

I don't understand why it is necessary to restrict the values of u so that each x-value in the domain of the integrand corresponds to exactly one value of u. I'm guessing the same concept applies when doing trignometric substitutions. Can someone please explain the whole restriction thing to me?

Thanks

Bored Of Fail · Jan 8, 2011

when you sub u= sqrt( x^2 +1 ) you are saying that u must be positive, the sqrt of a number must be positive

u^2 = x^2 +1 is pretty much the same substitution

with the substitution u^2 = x^2 +1 , if you had "x limits" which you had to change to "u limits" you would have to sub them into the formula and solve for u, and you would get two answers ( one positive and one negative ) when you take the sqrt

WEMG · Jan 8, 2011

Ok, but I've seen some worked examples done involving trig substitution and they don't mention any restrictions on domain in the working out. Do I lose any marks for not including it?

Bored Of Fail · Jan 8, 2011

WEMG said:
Ok, but I've seen some worked examples done involving trig substitution and they don't mention any restrictions on domain in the working out. Do I lose any marks for not including it?

yeh you dont have to worry about it at all, I have no idea why they really bother mentioning it.

It is a waste of time learning it

all you need to know are the main trig substitutions ( ie sin, sec and tan substitution ) and when to use them

Bored Of Fail · Jan 8, 2011

all the stuff about domain is just where the trig functions are monotonically increasing/decreasing ( im pretty sure, havent looked at it alot but thats my guess ) ( ie when an inverse exist ) and where they are one to one

WEMG · Jan 8, 2011

thanks

Drongoski · Jan 8, 2011

Using my weird, but equivalent, way for doing substitution-type problems, and ignoring all the considerations re domains etc:

$x^3dx = (x^2 +1 -1)xdx = (x^2 + 1 - 1) \dfrac 12 d(x^2 + 1)$

$\int \dfrac {x^3}{\sqrt {(x^2+1)} } dx$

$= \dfrac 12 \int [ (x^2+1)^{\frac 12} - (x^2+1)^{- \frac 12}] d(x^2+1)$

$= \dfrac 12 [\dfrac 23 (x^2+1)^ \frac 32 - 2 (x^2+1)^ \frac 12 ] + C$

$= \sqrt {(x^2+1)} (\frac13 (x^2+1) - 1) + C$

Bored Of Fail · Jan 8, 2011

^both confusing and stupid.

Drongoski · Jan 8, 2011

Agree.

Trebla · Jan 8, 2011

It would be better to state such restrictions for technical precision because otherwise you have to evaluate two different integrals with a positive and negative case (which both lead to the same answer). Ideally you should be doing both positive and negative cases if the change of variable is naturally unrestricted, but if you opt to do only one case make sure you state the assumption you made on the substitution.

If you decide to let u² = x²+1, then
u du = x dx

$\displaystyle\int\dfrac{x^3}{\sqrt{x^2+1}}\,dx=\displaystyle\int\dfrac{u(u^2-1)}{|u|}\,du\\\\$If $u > 0$ , then $|u| = u=+\sqrt{x^2+1}$ so integral becomes $\\\\\displaystyle\int\dfrac{u(u^2-1)}{u}\,du=\displaystyle\int(u^2-1)\,du\\\\=\dfrac{u^3}{3}-u+c\\\\=\dfrac{(x^2+1)^{\frac{3}{2}}}{3}-\sqrt{x^2+1}+c\\\\$If $u < 0$ , then $|u| = - u \Rightarrow u = - \sqrt{x^2+1}$ so integral becomes $\\\\\displaystyle\int\dfrac{u(u^2-1)}{-u}\,du=\displaystyle\int(1-u^2)\,du\\\\=u-\dfrac{u^3}{3}+c\\\\=-\sqrt{x^2+1}+\dfrac{(x^2+1)^{\frac{3}{2}}}{3}+c$

Notice both lead to the same answer.

Using substitution to reduce integrals to standard form (1 Viewer)

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