How do you evaluate the integral #int 1/(xsqrt(1-x^2))#?

Question

Charles Arocha · Accepted Answer

The answer is# =-1/2ln(sqrt(1-x^2)+1)+1/2ln(|sqrt(1-x^2)-1|)+C#

We perform this integral by substitution

Let #u=sqrt(1-x^2)#

#u^2=1-x^2#

#du=(-2xdx)/(2sqrt(1-x^2))=(-xdx)/(sqrt(1-x^2))#

Therefore,

#intdx/(xsqrt(1-x^2))=intsqrt(1-x^2)(du)/(-x^2sqrt(1-x^2))#

#=int-(du)/x^2#

#=int(du)/(u^2-1)#

Now, we perform a de composition into partial fractions

#1/(u^2-1)=1/((u+1)(u-1))=A/(u+1)+B/(u-1)#

#=(A(u-1)+B(u+1))/((u+1)(u-1))#

The denominators are the same, we can compare the numerators

#1=A(u-1)+B(u+1)#

Let #u=-1#, #=>#, #1=-2A#, #=>#, #A=-1/2#

Let #u=1#, #=>#, #1=2B#, #=>#, #B=1/2#

Therefore,

#1/(u^2-1)=(-1/2)/(u+1)+(1/2)/(u-1)#

So,

#int(du)/(u^2-1)=-1/2int(du)/(u+1)+1/2int(du)/(u-1)#

#=-1/2ln(u+1)+1/2ln(u-1)#

#=-1/2ln(sqrt(1-x^2)+1)+1/2ln(|sqrt(1-x^2)-1|)+C#

Adam Adkins · Answer

undefined To evaluate the integral $ \int \frac{1}{x\sqrt{1-x^2}} $, we can use trigonometric substitution. Let $ x = \sin(\theta) $, then $ dx = \cos(\theta) d\theta $ and $ \sqrt{1 - x^2} = \sqrt{1 - \sin^2(\theta)} = \sqrt{\cos^2(\theta)} = \cos(\theta) $.

After substitution, the integral becomes:

$$ \int \frac{1}{\sin(\theta) \cdot \cos(\theta)} \cdot \cos(\theta) \, d\theta $$

$$ = \int \frac{1}{\sin(\theta)} \, d\theta $$

$$ = \int \csc(\theta) \, d\theta $$

Now, this is a standard integral which integrates to $-\ln|\csc(\theta) + \cot(\theta)| + C$. However, since we are integrating with respect to $x$, we need to express the result in terms of $x$.

Recall that $x = \sin(\theta)$. Therefore, $ \csc(\theta) = \frac{1}{\sin(\theta)} = \frac{1}{x} $. Also, $ \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} = \frac{\sqrt{1 - x^2}}{x} $.

Thus, the integral becomes:

$$ = -\ln|\frac{1}{x} + \frac{\sqrt{1 - x^2}}{x}| + C $$

$$ = -\ln|\frac{\sqrt{1 - x^2} + 1}{x}| + C $$

This is the result after evaluating the integral $ \int \frac{1}{x\sqrt{1-x^2}} $.