Is $\int \frac{\sqrt{1-x^2}}{x}\;dx = \ln\lvert x\rvert + C$?

Question

I have the integral $$ \int \frac{\sqrt{1-x^2}}{x}\;dx $$ I do a trig. substitution $x = \sin(\theta)$. Then I get $\sqrt{1-x^2} = \cos(\theta)$. So $$ \int \frac{\sqrt{1-x^2}}{x}\;dx = \int \frac{\cos(\theta)}{\sin(\theta)}\; d\theta = \ln\lvert \sin(\theta)\rvert + C= \ln\lvert x\rvert + C. $$ But I am thinking that I made a mistake somewhere because the derivative of $\ln\lvert x\rvert$ is not $\frac{\sqrt{1-x^2}}{x}$.

Where is my mistake?

Answer 1

The omission is $dx=\cos(\theta)d\theta$.

Answer 2

$$\int\dfrac{\sqrt{1-x^2}}x\ dx=\int\dfrac{\cos^2\theta}{\sin\theta}d\theta =\int\dfrac{1-\sin^2\theta}{\sin\theta}d\theta=?$$

Answer 3

$$x=\sin\theta\implies dx=\cos(\theta)\,d\theta$$

$$\begin{align} \int{ \sqrt{1-x^2} \over x}\,dx &= \int{ \sqrt{1-\sin^2\theta} \over \sin\theta} \cos(\theta)\,d\theta \\ &= \int{ \sqrt{\cos^2\theta} \over \sin\theta } \cos(\theta)\,d\theta \\ &= \int{ \cos^2\theta \over \sin\theta}\,d\theta \\ &= \int{ -\sin^2(\theta) + 1 \over \sin\theta}\,d\theta \\ &= \int\bigl[ -\sin(\theta) + \csc(\theta)\bigr]\,d\theta \end{align}$$

You got it from there.

Answer 4

$$\displaystyle \int\dfrac{\sqrt{1-x^2}}x\ dx=\int\dfrac{\cos^2\theta}{\sin\theta}d\theta =\int(\frac{d\theta}{\sin\theta}-\sin\theta) d\theta \\ =\int(\frac{\sin\theta}{1-\cos^{2}\theta}-\sin\theta)d\theta$$