Integrate $\int\Big(\sqrt{4x^2+9-4x}\Big)^3\,dx$ by using trigonometric substitution.

Question

Hello I was wondering if I could have some assistance on integrating the following integral using trigonometric substitution. $$\int\Big(\sqrt{4x^2+9-4x}\Big)^3\,dx.$$

Answer 1

Note that $4x^2-4x+9=(2x-1)^2+8$.

So now put $2x-1=2\sqrt 2\tan\theta$

Then our integrand becomes $8\sqrt 2\sec^5\theta d\theta$

Can you do the integration now?

Answer 2

Hint

Completing the square $$4x^2-4x+9=4\left(x-\frac 12\right)^2+8$$ So, make the change of variable $$x=\frac 12+\sqrt 2 \sinh(y),\quad dx=\sqrt 2 \cosh(y)dy$$ and simplify. You should get to a quite simple expression.

I am sure that you can take from here.

Answer 3

$$4x^2+4x+9=(2x+1)^2+8\\2x+1:=\sqrt 8 \sinh t\Rightarrow I=32\int\cosh^4t\,\,dt=32\int\cosh^2t(1+\sinh^2t)\,\,dt=32\int{{(1+\cosh2t)}\over 2}+{1\over 8}(\cosh4t-1)\,\,dt={3\over 8}t+{1\over 4}\sinh 2t+{1\over 32}\sinh 4t$$