Question

Can you help me solve this integral and explain how it's done ??? 8x/(-4x^4+12x^2+27)^1/2 dx

Answer 1

\[\int\frac{8x}{\sqrt{-4x^4+12x^2+27}}~dx~~?\]
Complete the square in the denominator.
\[\begin{align*}
-4x^4+12x^2+27&=-4\left(x^4-3x^2-\frac{27}{4}\right)\\\\
&=-4\left(x^4-3x^2+\frac{9}{4}-\frac{36}{4}\right)\\\\
&=-4\left(\left(x^2-\frac{3}{2}\right)^2-\frac{36}{4}\right)\\\\
&=36-4\left(x^2-\frac{3}{2}\right)^2
\end{align*}\]
Use a substitution, \(u=x^2-\dfrac{3}{2}\), then \(\dfrac{1}{2}~du=x~dx\).
\[4\int\frac{du}{\sqrt{36-u^2}}\]
Make another substitution, \(u=6\sin t\), so that \(du=6\cos t~dt\).
\[24\int\frac{\cos t}{\sqrt{36-36\sin^2t}}~dt\\
4\int dt\]

Answer 2

sorry, ammmm why in the second substitution u= 6sinx & du = 6cosx dx i can't understand xD

Answer 3

For the \(u\)-integral
\[4\int\frac{du}{\sqrt{36-u^2}}\]
you need a substitution that can work with the otherwise difficult-to-integrate expression in the denominator. For expressions of this form, we use some useful trig identities.

Recall the Pythagorean identity:\[\sin^2x+\cos^2x=1\]
which gives \(1-\sin^2x=\cos^2x\). Note the similarity between the left side of this identity and the denominator - we have a difference of (not necessarily perfect) squares.

For this reason, we'll let \(u=6\sin t\), so that \(du=6\cos t~dt\). (Nothing fancy here, just taking thing the differentials.)

Then you substitute as you normally would:
\[\begin{align*}
4\int\frac{\color{blue}{du}}{\sqrt{36-\color{red}u^2}}&=4\int\frac{\color{blue}{6\cos t}}{\sqrt{36-\color{red}{(6\sin t^2)}}}~\color{blue}{dt}\\\\
&=24\int\frac{\cos t}{\sqrt{36-36\sin^2t}}~dt\\\\
&=\frac{24}{\sqrt{36}}\int\frac{\cos t}{\sqrt{1-\sin^2t}}~dt\\\\
&=\frac{24}{6}\int\frac{\cos t}{\sqrt{\cos^2t}}~dt\\\\
&=4\int\frac{\cos t}{\cos t}~dt\\\\
&=4\int dt
\end{align*}\]