Problem 2 Evaluate the following integrals... [FREE SOLUTION]

Chapter 1: Problem 2

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

Short Answer

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

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91影视

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

Short Answer

Step by step solution

Identify the suitable trigonometric substitution

Find the differential

Substitute and simplify the integrand

Evaluate the integral

Revert back to the original variable

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