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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 \(\sqrt{a^2+u^2}\) \(u=a\,\texttt{tan}(\theta)\)
* Note: The presence of the square-root function is not required to implement these techniques. 

Virtual Lessons

Need some additional help understanding how to apply this integration technique? Click Here to visit the virtual lesson section.

Practice Problems

Answer the following questions by using the integration technique known as trigonometric-substitution

  1. \(\displaystyle \int  \sqrt{1-9x^2} dx \)
  2. \(\displaystyle \int \sqrt{9-25x^2} dx \)
  3. \(\displaystyle \int \frac{\sqrt{1-x^2}}{x} dx \)
  4. \(\displaystyle \int \frac{x^2}{(1-4x^2)^{\frac{3}{2}}} dx \)
  5. \(\displaystyle \int \sqrt{x^2+4} dx \)
  1. \(\displaystyle \int \frac{dx}{x^2\sqrt{x^2+4}} \)
  2. \(\displaystyle \int^2_3 \frac{dx }{x^2-1} \)
  3. \(\displaystyle \int^4_3 \sqrt{x^2-4} dx \)
  4. \(\displaystyle \int \frac{dx }{\sqrt{x^2-9}}\; x<-3 \)
  5. \(\displaystyle \int^2_1 \frac{\texttt{e}^{3x}}{1-\texttt{e}^{2x}} dx \)
  1. \(\displaystyle \int^2_1 \frac{dx }{x\sqrt{64x^4-1}} \)
  2. \(\displaystyle \int \frac{dx }{2x^2+4x+4} \)
  3. \(\displaystyle \int^1_0 \frac{dx }{x^2-4x+8} \)

Suggestive Solution Guide

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 \(\displaystyle \int \sqrt{9-25x^2}{x} dx=\frac{3}{5}x\sqrt{9-25x^2} + \frac{6}{5}\arcsin\left(\frac{3}{5}x\right) + C\)