Question

Evaluate the following integrals : $$\int \frac{\mathrm{dx}}{\sqrt{1-\mathrm{x}^{2}}-1}$$

Short answer

Question: Evaluate the indefinite integral $$\int \frac{\mathrm{dx}}{\sqrt{1-x^2}-1}$$ Answer: $$\int \frac{\mathrm{dx}}{\sqrt{1-x^2}-1} = \ln |1 - \sqrt{1 - x^2}| + C$$

Step-by-step solution

Identify the suitable trigonometric substitution

We can rewrite the square root expression in the integrand as follows: $$\sqrt{1 - x^2}$$ The most suitable trigonometric substitution for this expression is \(x = \sin(\theta)\). This is because the expression will simplify with the identity \(\sin^2(\theta) + \cos^2(\theta) = 1\). Next, we need to find the differential \(\mathrm{d}x\).

Find the differential

To find \(\mathrm{d}x\), we differentiate the substitution expression with respect to \(\theta\): $$\frac{\mathrm{d}x}{\mathrm{d}\theta} = \frac{\mathrm{d}}{\mathrm{d}\theta}\sin(\theta) = \cos(\theta)$$ Then, multiply both sides by \(\mathrm{d}\theta\): $$\mathrm{d}x = \cos(\theta) \, \mathrm{d}\theta$$ With the substitution and differential in place, we are now ready to change the integral to the new variable \(\theta\).

Substitute and simplify the integrand

Substitute \(x = \sin(\theta)\) and \(\mathrm{d}x = \cos(\theta)\, \mathrm{d}\theta\) into the integral: $$\int \frac{\mathrm{d}x}{\sqrt{1-x^2}-1} = \int \frac{\cos(\theta) \, \mathrm{d}\theta}{\sqrt{1-\sin^2(\theta)}-1}$$ Now, simplify the integrand using the identity \(\sin^2(\theta) + \cos^2(\theta) = 1\): $$\int \frac{\cos(\theta) \, \mathrm{d}\theta}{\sqrt{1-\sin^2(\theta)}-1} = \int \frac{\cos(\theta) \, \mathrm{d}\theta}{\sqrt{\cos^2(\theta)}-1}$$ The square root of \(\cos^2(\theta)\) is \(\cos(\theta)\), so the integrand becomes: $$\int \frac{\cos(\theta) \, \mathrm{d}\theta}{\cos(\theta)-1}$$

Evaluate the integral

Notice that the integral can now be simplified even further: $$\int \frac{\cos(\theta) \, \mathrm{d}\theta}{\cos(\theta)-1} = \int \frac{\mathrm{d}\theta}{1-\cos(\theta)}$$ Now, use the substitution \(u = 1 - \cos(\theta)\) and \(\mathrm{d}u = \sin(\theta)\, \mathrm{d}\theta\): $$= \int \frac{\mathrm{d}u}{u}$$ This is a simple natural logarithmic integral, which evaluates to: $$\int \frac{\mathrm{d}u}{u} = \ln |u| + C$$

Revert back to the original variable

Now, we need to substitute back \(\theta\) and then \(x\). First, we substitute \(u = 1 - \cos(\theta)\): $$\ln |1 - \cos(\theta)| + C$$ Then, using the substitution \(x = \sin(\theta)\), we have \(\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - x^2}\): $$\ln |1 - \sqrt{1 - x^2}| + C$$ This is the final form of the indefinite integral: $$\int \frac{\mathrm{dx}}{\sqrt{1-x^2}-1} = \ln |1 - \sqrt{1 - x^2}| + C$$