Jtdylktuy

So I've looked up how to do this problem and when they complete the square it's using the $(frac{-b}{a})^2$ method where, in the denominator, they factor out the $4$, and then make:

$$x^2-x=x^2-x+{1over4}$$ and then

$$x^2-x+{1over4}=(x-frac{1}{2})^2$$

which I understand, but is not the first option that popped into my head.

What I don't understand is why all the tutorials I've looked up are not completing the square in the traditional method. Here's my work:

$intfrac{x^2}{sqrt{3+4x-4x^2}^3}dx$

denominator:

$$3+4x-4x^2$$
$$(-4x^2+4x-1)+4$$
$$-(4x^2-4x+1)+4$$
$$-(2x-1)^2+4$$

So the integral is now:

$$intfrac{x^2}{(4-u^2)^frac{3}{2}}$$

So I do the rest of the work by using trig substitution now:

$u=2x-1$

$x=frac{u+1}{2}$

Now I subbed in the trig identities:

$u=2sintheta$

$du=2costheta dtheta$

$$intfrac{(frac{u+1}{2})^22costheta}{sqrt{4-4sin^2theta}^3}dtheta$$

$$intfrac{frac{(2sintheta+1)^2}{4}2costheta}{sqrt{4cos^2theta}^3}dtheta$$

$$intfrac{frac{(2sintheta+1)^2}{4}2costheta}{(2costheta)^3}dtheta$$

$$intfrac{frac{(2sintheta+1)^2}{4}2costheta}{8cos^3theta}dtheta$$

$$intfrac{frac{(4sin^2theta+4sintheta+1)}{4}2costheta}{8cos^3theta}dtheta$$

$$intfrac{(frac{4sin^2theta}{4}+frac{4sintheta}{4}+frac{1}{4})2costheta}{8cos^3theta}dtheta$$

$$intfrac{(sin^2theta+sintheta+frac{1}{4})2costheta}{8cos^3theta}dtheta$$

$$intfrac{(sin^2theta+sintheta+frac{1}{4})}{8cos^3theta}*frac{2costheta}{1}dtheta$$

$$intfrac{(sin^2theta+sintheta+frac{1}{4})}{4cos^2theta}dtheta$$

$$intfrac{sin^2theta}{4cos^2theta}dtheta+intfrac{sintheta}{4cos^2theta}dtheta+intfrac{frac{1}{4}}{4cos^2theta}dtheta$$

$$frac{1}{4}int tan^2theta dtheta+frac{1}{4}intfrac{sintheta}{cos^2theta}dtheta+intfrac{1}{16cos^2theta}dtheta$$

For the second integral, I subbed $u=costheta$ and $-du=sintheta dtheta$

$$frac{1}{4}int sec^2theta-1dtheta-frac{1}{4}intfrac{du}{u^2}+frac{1}{16}int sec^2theta dtheta$$

$$[frac{tantheta}{4}-frac{theta}{4}]+[frac{1}{4costheta}]+[frac{tantheta}{16}]+C$$

On a right triangle where $sintheta=frac{2x-1}{2}$ and $costheta=frac{sqrt{4-(2x-1)^2}}{2}$, I subbed these values back in to make the integral in terms of $x$:

$$frac{2x-1}{4sqrt{4-(2x-1)^2}}-frac{arcsin(frac{2x-1}{2})}{4}+frac{1}{4(frac{sqrt{4-(2x-1)^2}}{2})}+frac{2x-1}{16sqrt{4-(2x-1)^2}}+C$$

This does not seem to be the right answer when I compare it to a calculator answer, so I'm guessing I either completed the square wrong or made an algebraic mistake? I've done this problem three or four times now and am getting a little frustrated over what I'm guessing is a easy fix. If anyone can help, I'd really appreciate it.

Except some minor mistakes, your solution is correct but you led to wrong answer.

Mistake 1

In the first integral after substituting $u=sin theta$ you used $dx=du$ while $dx={1over 2}du$.

Mistake 2

From $2$nd to $3$rd integral, you used $|cos theta|=cos theta$ while you should have dealt with $|cos theta|$ till the result.

Fixing up

Fixing these two mistakes, the final integral to be solved would become$$intfrac{sin^2theta}{8costheta|costheta|}dtheta+intfrac{sintheta}{8costheta|costheta|}dtheta+intfrac{1}{32costheta|costheta|}dtheta\={4+5sintheta-4thetacosthetaover 32|costheta|}+C$$

Find $I=intfrac{x^2}{(3+4x-4x^2)^{3/2}}dx$.

First 4 steps:

• Change $x=t+frac12$
  $$I=intfrac{t^2+t+frac14}{8(1-t^2)^{3/2}}dt\=frac18intfrac{t^2}{(1-t^2)^{3/2}}dt+
  frac18intfrac{t}{(1-t^2)^{3/2}}dt+frac1{32}intfrac{1}{(1-t^2)^{3/2}}dt$$



• $$intfrac{t}{(1-t^2)^{3/2}}dt=
  frac{1}{sqrt{1-{{t}^{2}}}}$$


• $$intfrac{t^2}{(1-t^2)^{3/2}}dt=frac{t}{sqrt{1-{{t}^{2}}}}-arcsin(t)$$


• $$intfrac{1}{(1-t^2)^{3/2}}dt=frac{t}{sqrt{1-{{t}^{2}}}}$$