Could someone explain to me how to solve this integration??
integrate $$\int\frac{dx}{\sqrt{(x^2-0.01)}}$$
The final answer is $$\ln({\sqrt{(x^2-0.01)}}+x)+C$$
How to get to the final answer?? Could someone show me the steps.
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Note that: $$\sqrt{x^{2}-0.01}=\sqrt{x^{2}-0.1^{2}}$$ Therefore, we can use the trigonometric substitution: $x=0.1\sec{\theta} \implies dx=0.1\sec{\theta}\tan{\theta}\:d\theta$ Therefore, our integral becomes, using the equality $\sec^{2}{x}-1=\tan^{2}{x}$: $$\int{\frac{dx}{\sqrt{x^{2}-0.01}}}=\int{\frac{\sec{\theta}\tan{\theta}}{\tan{\theta}}\:d\theta}=\int{\sec{\theta}\:d\theta}=\ln{\left|\sec{\theta}+\tan{\theta}\right|}+c$$ Now we back substitute, knowing that $\tan{\theta}=10\sqrt{x^{2}-0.01}$ and $\sec{\theta}=10x$, to get: $$\ln{|10(\sqrt{x^{2}-0.01}+x)|}+c$$ But the factor of 10 comes out of the logarithm as a constant, and so we can therefore write: $$\int{\frac{dx}{\sqrt{x^{2}-0.01}}}=\ln|\sqrt{x^{2}-0.01}+x|+c\qquad\text{Q.E.D}$$ |
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Remember when the integrand is as form $$R(\sqrt{ax^2+bx+c},x)$$ than if $a>0$ then you take a new substitution $\sqrt{ax^2+bx+c}=t+x\sqrt{a}$. Now here $a=1$ and you have $\sqrt{x^2-0.01}=t+x$ and $x=\frac{-t^2-0.01}{2t}$ and $dx=\frac{-2t^2+0.02}{4t^2}$. Can you do the rest?;-) |
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Hint: $x=0.1\cdot\sec(\theta)$ and $\mathrm dx=0.1 \tan(\theta)\sec(\theta) \mathrm d \theta$ |
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