Integration with logarithmic expression.

Question

I'm trying to solve an integration problem from the book which is the $$\int\frac{\sqrt{9-4x^2}}{x}dx$$ using trigonometric substitution. The answer from the book is $$3\ln\left|\frac{3-\sqrt{9-4x^2}}{x}\right|+\sqrt{9-4x^2}+C.$$ I have almost the same solution where there's a $$3\ln|\csc\theta-\cot\theta|+3\cos\theta+C.$$ The problem is when the substitution comes in. I end up having $$3\ln\frac{|3-\sqrt{9-x^2}|}{2x}+\sqrt{9-x^2}$$ and when I tried to simplify it further, it resulted to $$3\ln\left|3-\sqrt{9-x^2}\right|-3\ln|2x|+\sqrt{9-x^2}.$$ I hope you could help me to tell where i did wrong.

By the way, I set $a=3$ and $x=\frac{3}{2}\sin\theta$.

Answer 1

I tried to follow your steps. Here we are assuming that $0<x\leq 3/2$. Let $x=3/2\sin(t)$ then $$3\cos(t)=3\sqrt{1-(2x/3)^2}=\sqrt{9-4x^2}$$ and \begin{align*}\int\frac{\sqrt{9-4x^2}}{x}dx&=\int\frac{3\cos(t)}{3/2\sin(t)}d(3/2\sin(t))\\& =3\int\frac{\cos^2(t)}{\sin(t)}dt=3\int\frac{1-\sin^2(t)}{\sin(t)}dt\\&=3\int\frac{dt}{\sin(t)}-3\int \sin(t)dt \\&= 3\ln\frac{1-\cos(t)}{\sin(t)}+3\cos(t)+C\\ \\&= 3\ln\frac{3-3\cos(t)}{3\sin(t)}+3\cos(t)+C\\ &=3\ln\frac{3-\sqrt{9-4x^2}}{2x}+\sqrt{9-4x^2}+C\\ &=3\ln\frac{3-\sqrt{9-4x^2}}{x}+\sqrt{9-4x^2}+C'. \end{align*} where $C'=C-3\ln(2)$ is an arbitrary constant.

P.S. We have that for $\sin(t)>0$, \begin{align*} \int\frac{dt}{\sin(t)}&=\int\frac{1+\cos(t)-\cos(t)}{\sin(t)}\,dt\\ &=\int\frac{(1+\cos(t))(1-\cos(t))}{\sin(t)(1-\cos(t)}\,dt-\int\frac{\cos(t)}{\sin(t)}\,dt\\ &=\int\frac{\sin(t)}{1-\cos(t)}\,dt-\int\frac{\cos(t)}{\sin(t)}\,dt\\ &=\ln(1-\cos(t))-\ln(\sin(t))+C=\ln\left(\frac{1-\cos(t)}{\sin(t)}\right)+C \end{align*}

Answer 2

It is easy to mess up with the constants in such integrals. But we can temporarily ignore them because a simple scaling of the variable and of the integrand can bring us to

$$I:=\int\frac{\sqrt{1-t^2}}tdt.$$

This calls for the substitution $1-t^2=u^2$, giving $dt/t=-u\,du/2t^2$ and

$$I\propto\int\frac{u^2}{u^2-1}du=\int\frac{u^2-1+1}{u^2-1}du=u-\text{artanh u}=\sqrt{1-t^2}-\text{artanh}\sqrt{1-t^2}.$$

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Now back to the original problem, we now know that the solution will be of the form

$$\int\frac{\sqrt{9-4x^2}}xdx=\lambda\left(\sqrt{1-\frac49x^2}-\text{artanh}\sqrt{1-\frac49x^2}\right).$$

Differentiating, we get

$$\frac{\sqrt{9-4x^2}}x=-\lambda\frac{\dfrac49x}{\sqrt{1-\dfrac49x^2}}\left(1-\frac1{1-\left(1-\dfrac49x^2\right)}\right)=\lambda\frac{\sqrt{1-\dfrac49x^2}}x$$ and $\lambda=3$.

The final solution is

$$\sqrt{9-4x^2}-3\text{ artanh}\frac{\sqrt{9-4x^2}}3.$$

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Note:

We can use the following transformation

$$\text{artanh}\sqrt{1-t^2}=\log\sqrt{\frac{1+\sqrt{1-t^2}}{1-\sqrt{1-t^2}}}=\log\left|\frac{1+\sqrt{1-t^2}}t\right|.$$

Answer 3

Here's what I did. $$\int\frac{\sqrt{9-4x^2}}{x}dx \\ \begin{align} & a=3\\ & u=2x\\ \text{since} \quad& u=a\sin\theta \\ & 2x=3\sin\theta \\ & 2dx=3\cos\theta; \quad x=\frac{3}{2}\sin\theta; \quad \sin\theta=\frac{2x}{3}; \quad \theta= \sin^1\frac{2x}{3}\\ & dx = \frac{3}{2}\cos\theta \\ \\ \sqrt{a^2-u^2}& =\sqrt{9-(2x)^2} \\ & = \sqrt{9-(3sin\theta)^2} \\ & = \sqrt{9-(9sin^2\theta)}\\ & = 3\cos\theta \\ \\ \text{Going back to the problem.}\\ \int\frac{\sqrt{9-4x^2}}{x}dx &=\int\frac{3\cos\theta}{\frac{3}{2}\sin\theta}(\frac{3}{2}\cos\theta d\theta) \\ &=3\int\frac{\cos^2\theta}{\sin\theta} \\ &=3\int\frac{1}{\sin\theta}-3\int\sin\theta \\ &=3\ln(\csc\theta-\cot\theta)+3\cos\theta \\ &=3\ln(\frac{3}{2x}-\frac{\sqrt{9-4x^2}}{2x})+3\frac{\sqrt{9-4x^2}}{3} \\ &=3\ln(\frac{3}{2x}-\frac{\sqrt{9-4x^2}}{2x})+\sqrt{9-4x^2} \\ &=3\ln(\frac{3-\sqrt{9-4x^2}}{2x})+\sqrt{9-4x^2} \\ &=3(\ln(3-\sqrt{9-4x^2}-\ln(2x))+\sqrt{9-4x^2} \\ &=3\ln(3-\sqrt{9-4x^2}-3\ln(2x)+\sqrt{9-4x^2} \\ \end{align} $$