Solutions to Problem Set 4

Math 211-03

9-11-2017

[Trig substitution]

1. Compute $\displaystyle \int
   \dfrac{1}{\sqrt{9 - x^2}}\,dx$ .

You don't need to do a trig substitution, since this is a standard inverse sine form.

$$\int \dfrac{1}{\sqrt{9 - x^2}}\,dx = \sin^{-1} \dfrac{x}{3} + c.$$

However, you could do this using trig substitution, using $x = 3 \sin \theta$ .


2. Compute $\displaystyle \int
   \dfrac{1}{(4 + x^2)^{5/2}}\,dx$ .

$$\int \dfrac{1}{(4 + x^2)^{5/2}}\,dx = \int \dfrac{1}{(4 + (2 \tan \theta)^2)^{5/2}} 2 (\sec \theta)^2\,d\theta = \int \dfrac{1}{(4 + 4 (\tan \theta)^2)^{5/2}} 2 (\sec \theta)^2\,d\theta =$$

\hfil\raise0.5in\hbox{$\left[x = 2
   \tan \theta, \quad dx = 2 (\sec \theta)^2\,d\theta\right]$ } \hskip0.5in \hbox{\epsfysize=1in \epsffile{tangent-triangle2.eps}}\hfil

$$\int \dfrac{1}{4 (\sec \theta)^2)^{5/2}} 2 (\sec \theta)^2\,d\theta = \int \dfrac{1}{32 (\sec \theta)^5} 2 (\sec \theta)^2\,d\theta =$$

$$\dfrac{1}{16} \int \dfrac{1}{(\sec \theta)^3}\,d\theta = \dfrac{1}{16} \int (\cos \theta)^3\,d\theta = \dfrac{1}{16} \int (\cos \theta)^2 (\cos \theta)\,d\theta = \dfrac{1}{16} \int [1 - (\sin \theta)^2] (\cos \theta)\,d\theta =$$

$$\left[u = \sin \theta, \quad du = \cos \theta\,d\theta, \quad d\theta = \dfrac{du}{\cos \theta}\right]$$

$$\dfrac{1}{16} \int (1 - u^2) (\cos \theta) \cdot \dfrac{du}{\cos \theta} = \dfrac{1}{16} \int (1 - u^2)\,du = \dfrac{1}{16} \left(u - \dfrac{1}{3} u^3\right) + c =$$

$$\dfrac{1}{16} \left(\sin \theta - \dfrac{1}{3} (\sin \theta)^3\right) + c = \dfrac{1}{16} \left(\dfrac{x}{\sqrt{4 + x^2}} - \dfrac{1}{3} \dfrac{x^3}{(4 + x^2)^{3/2}}\right) + c.\quad\halmos$$


3. Compute $\displaystyle \int
   \dfrac{1}{x^2 \sqrt{1 - x^2}}\,dx$ .

$$\int \dfrac{1}{x^2 \sqrt{1 - x^2}}\,dx = \int \dfrac{1}{(\sin \theta)^2 \sqrt{1 - (\sin \theta)^2}}(\cos \theta\,d\theta) = \int \dfrac{1}{(\sin \theta)^2 \sqrt{(\cos \theta)^2}}(\cos \theta\,d\theta) =$$

$$\hfil\raise0.5in\hbox{$\left[x = \sin \theta, \quad dx = \cos \theta\,d\theta\right]$} \hskip0.5in \hbox{\epsfysize=1in \epsffile{sine-triangle1.eps}}\hfil$$

$$\int \dfrac{1}{(\sin \theta)^2 \cdot \cos \theta} (\cos \theta\,d\theta) = \int \dfrac{1}{(\sin \theta)^2}\,d\theta = \int (\csc \theta)^2\,d\theta = -\cot \theta + C = - \dfrac{\sqrt{1 - x^2}}{x} + c.\quad\halmos$$


4. Compute $\displaystyle \int
   \sqrt{4 - x^2}\,dx$ .

$$\int \sqrt{4 - x^2}\,dx = \int \sqrt{4 - 4 (\sin \theta)^2} (2 \cos \theta\,d\theta) = \int \sqrt{4 (\cos \theta)^2}(2 \cos \theta\,d\theta) =$$

$$\hfil\raise0.5in\hbox{$\left[x = 2 \sin \theta, \quad dx = 2 \cos \theta\,d\theta\right]$} \hskip0.5in \hbox{\epsfysize=1in \epsffile{sine-triangle2.eps}}\hfil$$

$$\int (2 \cos \theta)(2 \cos \theta\,d\theta) = 4 \int (\cos \theta)^2\,d\theta = 4 \int \dfrac{1}{2}(1 + \cos 2 \theta)\,d\theta = 2 \left(\theta + \dfrac{1}{2} \sin 2 \theta\right) + c =$$

$$2 \left(\theta + \dfrac{1}{2} \cdot 2 \sin \theta \cos \theta\right) + c = 2 \left(\sin^{-1} \dfrac{x}{2} + \dfrac{x \sqrt{4 - x^2}}{4}\right) + c. \quad\halmos$$


5. Compute $\displaystyle \int x
   \sqrt{4 - x^2}\,dx$ .

While you could do this using trig substitution, it's unnecessarily complicated. A simple substitution is better.

$$\int x \sqrt{4 - x^2}\,dx = \int x \sqrt{u} \cdot \dfrac{du}{-2 x} = -\dfrac{1}{2} \int u^{1/2}\,du = -\dfrac{1}{2} \cdot \dfrac{2}{3} u^{3/2} + c = -\dfrac{1}{3} (4 - x^2)^{3/2} + c.$$

$$\left[u = 4 - x^2, \quad du = -2 x\,dx, \quad dx = \dfrac{du}{-2 x}\right]\quad\halmos$$


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