# Not getting the right answer with alternate completing the square method on $\int\frac{x^2}{\sqrt{3+4x-4x^2}^3}dx$

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:

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

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

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:

$$\int\frac{x^2}{\sqrt{3+4x-4x^2}^3}dx$$

denominator:

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

So the integral is now:

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

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:

$$u=2\sin\theta$$

$$du=2\cos\theta d\theta$$

$$\int\frac{(\frac{u+1}{2})^22\cos\theta}{\sqrt{4-4\sin^2\theta}^3}d\theta$$

$$\int\frac{\frac{(2\sin\theta+1)^2}{4}2\cos\theta}{\sqrt{4\cos^2\theta}^3}d\theta$$

$$\int\frac{\frac{(2\sin\theta+1)^2}{4}2\cos\theta}{(2\cos\theta)^3}d\theta$$

$$\int\frac{\frac{(2\sin\theta+1)^2}{4}2\cos\theta}{8\cos^3\theta}d\theta$$

$$\int\frac{\frac{(4\sin^2\theta+4\sin\theta+1)}{4}2\cos\theta}{8\cos^3\theta}d\theta$$

$$\int\frac{(\frac{4\sin^2\theta}{4}+\frac{4\sin\theta}{4}+\frac{1}{4})2\cos\theta}{8\cos^3\theta}d\theta$$

$$\int\frac{(\sin^2\theta+\sin\theta+\frac{1}{4})2\cos\theta}{8\cos^3\theta}d\theta$$

$$\int\frac{(\sin^2\theta+\sin\theta+\frac{1}{4})}{8\cos^3\theta}*\frac{2\cos\theta}{1}d\theta$$

$$\int\frac{(\sin^2\theta+\sin\theta+\frac{1}{4})}{4\cos^2\theta}d\theta$$

$$\int\frac{\sin^2\theta}{4\cos^2\theta}d\theta+\int\frac{\sin\theta}{4\cos^2\theta}d\theta+\int\frac{\frac{1}{4}}{4\cos^2\theta}d\theta$$

$$\frac{1}{4}\int \tan^2\theta d\theta+\frac{1}{4}\int\frac{\sin\theta}{\cos^2\theta}d\theta+\int\frac{1}{16\cos^2\theta}d\theta$$

For the second integral, I subbed $$u=\cos\theta$$ and $$-du=\sin\theta d\theta$$

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

$$[\frac{\tan\theta}{4}-\frac{\theta}{4}]+[\frac{1}{4\cos\theta}]+[\frac{\tan\theta}{16}]+C$$

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

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

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 $$dx={1\over 2}du$$.

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$$\int\frac{\sin^2\theta}{8\cos\theta|\cos\theta|}d\theta+\int\frac{\sin\theta}{8\cos\theta|\cos\theta|}d\theta+\int\frac{1}{32\cos\theta|\cos\theta|}d\theta\\={4+5\sin\theta-4\theta\cos\theta\over 32|\cos\theta|}+C$$

• So my $du$ is incorrect because where $u=2x-1$, $du=2dx$, thus $dx=\frac{1}{2}du$? I think I understand that, but doesn't $(4cos^2\theta)^\frac{3}{2}$ turn into $$\sqrt{4cos^2\theta}\sqrt{4cos^2\theta}\sqrt{4cos^2\theta}$$ which is equivalent to $$(2cos\theta)(2cos\theta)(2cos\theta)$$ so where do the absolute value signs come from? Unless that's just some rule I'm not aware of. – JSmith Jan 9 '19 at 19:04
• Note that $$\sqrt {x^2}=|x|\ne x$$for $x\in \Bbb R$ for example $$\sqrt{(-1)^2}=\sqrt{1^2}=|-1|=|1|=1$$ – Mostafa Ayaz Jan 9 '19 at 19:50
• So, just to make sure my algebra is right after re-doing this, the denominator is: $$\sqrt{4cos^2\theta}^3$$ $$(2|cos\theta|)^3$$ $$(2|cos\theta|)(2|cos\theta|)(2|cos\theta|)$$ So $(|cos\theta|)(|cos\theta|)=cos^2\theta$ because the squared part makes the absolute values negligent, leaving: $$8cos^2\theta|cos\theta|$$ And the $cos\theta$ up top makes: $$8cos\theta|cos\theta|$$ And that's how you got your denominator in your final answer in terms of $\theta$? – JSmith Jan 9 '19 at 21:35
• Yes you got it. That's accurate... – Mostafa Ayaz Jan 10 '19 at 10:11

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

First 4 steps:

• Change $$x=t+\frac12$$ $$I=\int\frac{t^2+t+\frac14}{8(1-t^2)^{3/2}}dt\\=\frac18\int\frac{t^2}{(1-t^2)^{3/2}}dt+ \frac18\int\frac{t}{(1-t^2)^{3/2}}dt+\frac1{32}\int\frac{1}{(1-t^2)^{3/2}}dt$$

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