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This integration technique is particularly useful whenever either a sum-of-square or a difference-of-squares is found within the integrand. This approach is typically implemented to help reduce the complexity found among square-root functions.
| Difference of Squares | ||
| Substitution | Expression | Protocols |
| Sine | \(\sqrt{a^2-u^2}\) | \(u=a\,\texttt{sin}(\theta)\) |
| Secant | \(\sqrt{u^2-a^2}\) | \(u=a\,\texttt{sec}(\theta)\) |
| Sum of Squares | ||
| Substitution | Expression | Protocols |
| Tangent | \(\sqrt{a^2+u^2}\) | \(u=a\,\texttt{tan}(\theta)\) |
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Answer the following questions by using the integration technique known as trigonometric-substitution.
We begin by making the substitution \(\displaystyle 3x=\sin\theta\). Then, we have \(\displaystyle 3dx=\cos\theta d\theta\) and \(\displaystyle\sqrt{1-9x^2}=\sqrt{1-\sin^2\theta}=\cos\theta\). Substituting these expressions into the integral, we get
\(\begin{aligned} \int \sqrt{1-9x^2}{x} dx &= \int \frac{1}{3}\sin\theta(\cos\theta)\left(\frac{1}{3}\cos\theta d\theta\right) \\ &= \frac{1}{9}\int \sin\theta\cos^2\theta d\theta. \end{aligned}\)Next, we make the substitution \(u=\cos\theta\). Then, we have \(du=-\sin\theta d\theta\), so \(\sin\theta=-\sqrt{1-u^2}\). Substituting these expressions into the integral, we get
\(\begin{align} \frac{1}{9}\int \sin\theta\cos^2\theta d\theta &= \frac{1}{9}\int (-u^2+1)du\\ &= \frac{1}{9}\left[-\frac{u^3}{3}+u\right]+C \\ &= -\frac{1}{27}(1-3\cos^2\theta)\sqrt{1-9x^2}+C \\ &= -\frac{1}{27}(1-3x^2)\sqrt{1-9x^2}+C. \end{align}\)
Therefore, the solution to \(\displaystyle \int \sqrt{1-9x^2}{x} dx=-\frac{1}{27}(1-3x^2)\sqrt{1-9x^2}+C\), where \(C\) is the constant of integration.
We start by making the substitution:
\(5x=3\sin\theta\)Then, we have:
\( \sqrt{9-25x^2} = \sqrt{9-25\left(\frac{3}{5}\sin\theta\right)^2} = \sqrt{9-9\sin^2\theta} = 3\cos\theta\)Also, we have:
\(\begin{aligned}5 dx &= 3\cos\theta d\theta \end{aligned}\)Substituting these expressions into the integral, we get:
\(\begin{aligned} \int \sqrt{9-25x^2} dx &= \int 3\cos^2\theta\cdot\frac{3}{5}\cos\theta,d\theta \\ &= \frac{9}{5}\int\cos^3\theta,d\theta \end{aligned}\)To evaluate this integral, we use the reduction formula:
\(\int \cos^n\theta,d\theta = \frac{1}{n}\cos^{n-1}\theta\sin\theta + \frac{n-1}{n}\int \cos^{n-2}\theta d\theta\)Using this formula with \(n=3\), we get:
\(\begin{aligned} \int \cos^3\theta,d\theta &= \frac{1}{3}\cos^2\theta\sin\theta + \frac{2}{3}\int \cos\theta,d\theta \\ &= \frac{1}{3}\cos^2\theta\sin\theta + \frac{2}{3}\sin\theta + C \end{aligned}\)where \(C\) is the constant of integration.
Substituting back for \(\theta\) and simplifying, we get:
\(\begin{aligned} \int \sqrt{9-25x^2},dx &= \frac{9}{5}\int\cos^3\theta d\theta \\ &= \frac{9}{5}\left(\frac{1}{3}\cos^2\theta\sin\theta + \frac{2}{3}\sin\theta\right) + C \\ &= \frac{3}{5}\sin\theta\cos^2\theta + \frac{6}{5}\sin\theta + C \\ &= \frac{3}{5}x\sqrt{9-25x^2} + \frac{6}{5}\arcsin\left(\frac{3}{5}x\right) + C \end{aligned}\)Therefore, the solution to \(\displaystyle \int \sqrt{9-25x^2}{x} dx=\frac{3}{5}x\sqrt{9-25x^2} + \frac{6}{5}\arcsin\left(\frac{3}{5}x\right) + C\)
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