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I'm supposed to use a table of integrals to solve the below equation:

$$\int y \sqrt{6+12y-36y^2}dy$$

I'm having trouble identifying the form to use because I guess my weakness in algebra shows in my inability to compete the square of the square root function.

Here is how I tried to complete the square (first rewriting the equation and factoring out a -12):

$$6-12(-y+3y^2)$$

And I assume the next step is dividing my $b$ by $2$ and then squaring the number which would give:

$$6-12(\frac{1}{4}-\frac{1}{2}y+3y^2)$$

But I don't think this is right and I don't know how to get it in the form of $\sqrt{a^2-u^2}$

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  • $\begingroup$ integral seems very pointless because missing $dy$, sir $\endgroup$ – Madrit Zhaku Oct 19 '13 at 18:12
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As $6+12y-36y^2=7-(6y-1)^2,$

Put $6y-1=\sqrt7\sin\theta$

$$\int y\sqrt{6+12y-36y^2}dy=\frac{\sqrt7\sin\theta+1}6\sqrt7\cos\theta\frac{\sqrt7\cos\theta}6 d\theta$$

$$=\frac7{36}\int(\sqrt7\sin\theta+1)\cos^2\theta d\theta$$

$$=\frac7{36}\sqrt7\int \sin\theta \cos^2\theta d\theta+\frac7{36}\int \cos^2\theta d\theta$$

For the first part, put $\cos\theta=u$

For the second, $$\int \cos^2\theta d\theta=\frac{(1+\cos2\theta)}2 d\theta=\frac{\theta}2+\frac{\sin2\theta}4$$

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  • $\begingroup$ Thanks, but I don't understand how you got $7-(6y-1)^2$ from $6+12y-36y^2$ $\endgroup$ – hax0r_n_code Oct 19 '13 at 18:24
  • $\begingroup$ @inquisitor, $$6+12y-36y^2=6-(36y^2-12y)$$ $$=6-\{(6y)^2-2\cdot6y\cdot1+1^2\}+1=7-(6y-1)^2$$ $\endgroup$ – lab bhattacharjee Oct 19 '13 at 18:26

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