Trigonometric Substitution

Tags

Calculus

Cegep/2

Word count

697 words

Reading time

5 minutes

Method of reversing applications of Pythagorean Theorem to find the antiderivative:

• $a^2-x^2$:
  • Let $x=a\sin \theta$, then $a^2{\cos }^2\theta$
  • Let $x=a\cos \theta$, then $a^2{\sin }^2\theta$
• $a^2+x^2$:
  • Let $x=a\tan \theta$, then $a^2{\sec }^2\theta$
  • Let $x=a\cot \theta$, then $a^2{\csc }^2\theta$
• $x^2-a^2$:
  • Let $x=a\sec \theta$, then $a^2{\tan }^2\theta$
  • Let $x=a\csc \theta$, then $a^2{\cot }^2\theta$

Proof

$\sqrt{a^2-x^2}$ for $a>0$

Let $x=a\sin \theta$, then $\theta =\arcsin \frac{x}{a}$.

$$\begin{array}{rl}\sqrt{a^2-x^2}& =\sqrt{a^2-(a\sin \theta {)}^2}\\ & =\sqrt{a^2-a^2{\sin }^2\theta }\\ & =\sqrt{a^2(1-{\sin }^2\theta )}\\ & =\sqrt{a^2{\cos }^2\theta }\end{array}$$

Because $\cos \theta \ge 0$ for $\theta \in [-\frac{\pi }{2},\frac{\pi }{2}]$,

$$=a\cos \theta$$

$\sqrt{a^2+x^2}$ for $a>0$

Let $x=a\tan \theta$, then $\theta =\arctan \frac{x}{a}$.

$$\begin{array}{rl}\sqrt{a^2+x^2}& =\sqrt{a^2+(a\tan \theta {)}^2}\\ & =\sqrt{a^2+a^2{\tan }^2\theta }\\ & =\sqrt{a^2(1+{\tan }^2\theta )}\\ & =\sqrt{a^2{\sec }^2\theta }\\ & =a\sec \theta \end{array}$$

$\sqrt{x^2-a^2}$ for $a>0$

Let $x=a\sec \theta$, then $\theta =\mathrm{arcsec}\frac{x}{a}$.

$$\begin{array}{rl}\sqrt{x^2-a^2}& =\sqrt{(a\sec \theta {)}^2-a^2}\\ & =\sqrt{a^2{\sec }^2\theta -a^2}\\ & =\sqrt{a^2({\sec }^2\theta -1)}\\ & =\sqrt{a^2{\tan }^2\theta }\\ & =a\tan \theta \end{array}$$

Examples

Evaluate the following integrals.

$\int \frac{1}{(x^2+9{)}^{\frac{3}{2}}}dx$

Let $x=3\tan \theta$, then $dx=3{\sec }^2\theta d\theta$.
By trig substitution,

$$\begin{array}{rl}\int \frac{1}{(x^2+9{)}^{\frac{3}{2}}}dx& =\int \frac{1}{{\sqrt{x^2+9}}^3}dx\\ & =\int \frac{3{\sec }^2\theta }{(3\sec \theta {)}^3}d\theta \\ & =\frac{1}{9}\int \frac{1}{\sec \theta }d\theta \\ & =\frac{1}{9}\int \cos \theta d\theta \\ & =\frac{1}{9}\sin \theta +C\\ & =\frac{1}{9}\left(\frac{x}{\sqrt{x^2+9}}\right)+C\end{array}$$

$\int \frac{\sqrt{x^2-2}}{x^3}dx$

Let $x=\sqrt{2}\sec \theta$, then $dx=\sqrt{2}\sec \theta \tan \theta d\theta$.
By trig substitution,

$$\begin{array}{rl}\int \frac{\sqrt{x^2-2}}{x^3}dx& =\int \frac{\sqrt{2}\tan \theta }{(\sqrt{2}\sec \theta {)}^3}\sqrt{2}\sec \theta \tan \theta d\theta \\ & =\frac{1}{\sqrt{2}}\int \frac{{\tan }^2\theta }{{\sec }^2\theta }d\theta \\ & =\frac{1}{\sqrt{2}}\int {\sin }^2\theta d\theta \\ & =\frac{1}{\sqrt{2}}\cdot \frac{1}{2}\int (1-\cos 2\theta )d\theta \\ & =\frac{1}{2\sqrt{2}}(\theta -\frac{\sin 2\theta }{2})+C\\ & =\frac{1}{2\sqrt{2}}\sqrt{\mathrm{arcsec}\frac{x}{\sqrt{2}}-\sin \theta \cos \theta }+C\\ & =\frac{1}{2\sqrt{2}}(\mathrm{arcsec}\frac{x}{\sqrt{2}}-\left(\frac{\sqrt{x^2-2}}{x}\right)\left(\frac{\sqrt{2}}{x}\right))+C\end{array}$$

$\int \sqrt{16-x^2}dx$

Let $x=4\sin \theta$, then $dx=4\cos \theta d\theta$.
By trig substitution,

$$\begin{array}{rl}\int \sqrt{16-x^2}dx& =\int 4\cos \theta \cdot 4\cos \theta d\theta \\ & =8\int (1+\cos 2\theta )d\theta \\ & =8(\theta +\frac{\sin 2\theta }{2})+C\\ & =8(\arcsin \frac{x}{4}+\left(\frac{x}{4}\right)\left(\frac{\sqrt{16-x^2}}{4}\right))+C\end{array}$$

$\int \frac{1}{x^2\sqrt{5+x^2}}dx$

Let $x=\sqrt{5}\tan \theta$, then $dx=\sqrt{5}{\sec }^2\theta d\theta$.
By trig substitution,

$$\begin{array}{rl}\int \frac{1}{x^2\sqrt{5+x^2}}dx& =\int \frac{1}{5{\tan }^2\theta \sqrt{5}\sec \theta }d\theta \\ & =\frac{1}{5}\int \frac{\sec \theta }{{\tan }^2\theta }d\theta \\ & =\frac{1}{5}\int \frac{\cos \theta }{{\sin }^2\theta }d\theta \end{array}$$

Let $u=\sin \theta$, then $du=\cos \theta d\theta$.
By substitution,

$$\begin{array}{rl}& =\frac{1}{5}\int \frac{1}{u^2}du\\ & =-\frac{1}{5}\left(\frac{1}{\sin \theta }\right)+C\\ & =-\frac{1}{5}\left(\frac{\sqrt{x^2+5}}{x}\right)+C\end{array}$$

$\int \frac{x}{\sqrt{x^2-7}}\mathrm{d}x$

Let $x=\sqrt{7}\sec \theta$, then $dx=\sqrt{7}\tan \theta \sec \theta d\theta$.
By trig substitution,

$$\begin{array}{rl}\int \frac{x}{\sqrt{x^2-7}}\mathrm{d}x& =\int \frac{\sqrt{7}\sec \theta }{\sqrt{7}\tan \theta }\sqrt{7}\tan \theta \sec \theta \mathrm{d}\theta \\ & =\sqrt{7}\int {\sec }^2\theta \mathrm{d}\theta \\ & =\sqrt{7}\tan \theta +C\\ & =\sqrt{x^2-7}+C\end{array}$$

$\int \frac{\ln (\arcsin t)}{\sqrt{1-t^2}}\mathrm{d}t$

Let $t=\sin \theta ⟹dt=\cos \theta d\theta$.

$$\begin{array}{rl}\int \frac{\ln (\arcsin t)}{\sqrt{1-t^2}}\mathrm{d}t& =\int \frac{\ln (\arcsin (\sin \theta ))}{\cos \theta }\cos \theta \mathrm{d}\theta \\ & =\int \ln \theta \mathrm{d}\theta \\ & =\theta ln\theta -\theta +C\\ & =\arcsin t\cdot \ln (\arcsin t)-\arcsin t+C\end{array}$$

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Ajitani Hifumi

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